I’ve always confused “permutation” and “combination” — which one’s which?

Here’s an easy way to remember: **permutation sounds complicated**, doesn’t it? And it is. With permutations, every little detail matters. Alice, Bob and Charlie is different from Charlie, Bob and Alice (insert your friends’ names here).

Combinations, on the other hand, are pretty easy going. The details don’t matter. Alice, Bob and Charlie is the same as Charlie, Bob and Alice.

**Permutations are for lists (order matters) and combinations are for groups (order doesn’t matter).**

A joke: A "combination lock" should really be called a "permutation lock". The order you put the numbers in matters. (A true "combination lock" would accept both 10-17-23 and 23-17-10 as correct.)

## Permutations: The hairy details

Let’s start with permutations, or **all possible ways** of doing something. We’re using the fancy-pants term “permutation”, so we’re going to care about every last detail, including the order of each item. Let’s say we have 8 people:

```
1: Alice
2: Bob
3: Charlie
4: David
5: Eve
6: Frank
7: George
8: Horatio
```

How many ways can we award a 1st, 2nd and 3rd place prize among eight contestants? (Gold / Silver / Bronze)

We’re going to use permutations since the order we hand out these medals matters. Here’s how it breaks down:

- Gold medal: 8 choices: A B C D E F G H (Clever how I made the names match up with letters, eh?). Let’s say A wins the Gold.
- Silver medal: 7 choices: B C D E F G H. Let’s say B wins the silver.
- Bronze medal: 6 choices: C D E F G H. Let’s say… C wins the bronze.

We picked certain people to win, but the details don’t matter: we had 8 choices at first, then 7, then 6. The total number of options was 8 · 7 · 6 = 336.

Let’s look at the details. We had to order 3 people out of 8. To do this, we started with all options (8) then took them away one at a time (7, then 6) until we ran out of medals.

We know the factorial is:

Unfortunately, that does too much! We only want 8 · 7 · 6. How can we “stop” the factorial at 5?

This is where permutations get cool: notice how we want to get rid of 5 · 4 · 3 · 2 · 1. What’s another name for this? 5 factorial!

So, if we do 8!/5! we get:

And why did we use the number 5? Because it was left over after we picked 3 medals from 8. So, a better way to write this would be:

where 8!/(8-3)! is just a fancy way of saying “Use the first 3 numbers of 8!”. If we have **n** items total and want to pick **k** in a certain order, we get:

And this is the fancy permutation formula: You have **n** items and want to find the number of ways **k** items can be ordered:

## Combinations, Ho!

Combinations are easy going. Order doesn’t matter. You can mix it up and it looks the same. Let’s say I’m a cheapskate and can’t afford separate Gold, Silver and Bronze medals. In fact, I can only afford empty tin cans.

How many ways can I give 3 tin cans to 8 people?

Well, in this case, the order we pick people doesn’t matter. If I give a can to Alice, Bob and then Charlie, it’s the same as giving to Charlie, Alice and then Bob. Either way, they’re equally disappointed.

This raises an interesting point — we’ve got some redundancies here. Alice Bob Charlie = Charlie Bob Alice. For a moment, let’s just figure out how many ways we can rearrange 3 people.

Well, we have 3 choices for the first person, 2 for the second, and only 1 for the last. So we have 3 · 2 · 1 ways to re-arrange 3 people.

Wait a minute… this is looking a bit like a permutation! You tricked me!

Indeed I did. If you have N people and you want to know how many arrangements there are for **all** of them, it’s just N factorial or N!

So, if we have 3 tin cans to give away, there are 3! or 6 variations for every choice we pick. If we want to figure out how many combinations we have, we just **create all the permutations and divide by all the redundancies**. In our case, we get 336 permutations (from above), and we divide by the 6 redundancies for each permutation and get 336/6 = 56.

The general formula is

which means “Find all the ways to pick k people from n, and divide by the k! variants”. Writing this out, we get our **combination formula**, or the number of ways to combine k items from a set of n:

Sometimes C(n,k) is written as:

which is the the binomial coefficient.

## A few examples

Here’s a few examples of combinations (order doesn’t matter) from permutations (order matters).

Combination: Picking a team of 3 people from a group of 10. C(10,3) = 10!/(7! · 3!) = 10 · 9 · 8 / (3 · 2 · 1) = 120.

Permutation: Picking a President, VP and Waterboy from a group of 10. P(10,3) = 10!/7! = 10 · 9 · 8 = 720.

Combination: Choosing 3 desserts from a menu of 10. C(10,3) = 120.

Permutation: Listing your 3 favorite desserts, in order, from a menu of 10. P(10,3) = 720.

Don’t memorize the formulas, understand why they work. **Combinations sound simpler than permutations, and they are. You have fewer combinations than permutations.**

Understanding the Birthday Paradox | BetterExplainedApril 26, 2007 at 12:33 am[…] (Brush up on combinations and permuations if you like). […]

jameswMarch 22, 2017 at 9:15 amhow to do 6c3

LeahMarch 24, 2017 at 6:44 pmin your calculation, press 10 and press C (Combination) of and then press 3

Rukhsana shaikhSeptember 27, 2017 at 7:12 pm120

Dylan BMay 22, 2017 at 11:11 am6!/((6-3)!*3!)

Rukhsana shaikhSeptember 27, 2017 at 7:12 pm20

David YuJune 4, 2017 at 10:57 amit’s 6p3/3p3 or (6*5*4)/3*2*1) which = 20.

Rukhsana shaikhSeptember 27, 2017 at 7:14 pm20and 3

Ansn2[fJune 11, 2017 at 4:25 pmIt is 6•5•4/3•2•=20

Pavanta.MJune 21, 2017 at 6:18 pm6*5*4*3*2*1/6-3!*3!=6*5*4*3*2*1/3*2*1*3*2*1*=720/36=20

GazaliJuly 15, 2017 at 11:15 pm6c3 can be written as;

(6!)/(6-3)!(3)!

= 6*5*4*3!/3!*3*2*1 (3! can cancel out), hence you have

=120/6

=20, I hope that answers your question

SakshiJuly 31, 2017 at 11:10 am6!/3!*3!

CharlieSeptember 3, 2017 at 12:03 am6c3 is 6!/3!3!

6x5x4x3x2x1/3x2x1x3x2x1

=20

AkashSeptember 7, 2017 at 2:45 am6c3=6p3/3!

=6!/(6-3)!3!

=(6*5*4*3!)/(3!*3!)

=(6*5*4)/(3*2)

=20

SubhiSeptember 11, 2017 at 10:16 am6!/3!(6-3)!

=6!/3!×3!

=6×5×4×3!/3!×3!

=6×5×4/3!

=6×5×4/3×2×1

Simplify then u will get..

=20 ans

Rukhsana shaikhSeptember 27, 2017 at 7:11 pm20

AvishNovember 19, 2017 at 12:49 am6c3= 6!/3!(6-3)! => 6!/3!(3!) => 20

andrewApril 20, 2017 at 9:41 amcan anyone show me how many combinations there are using only 1, 2, and 3 across and 1 to 6 down.and 3 down, in other words its a 3 x 6 grid. Please do not give me the formula, I just need to see something like this, 1x2x3-1x3x2-2x3x1-2x1x3-3x1x2-3x2xhope it makes sense

LisaMay 8, 2017 at 1:16 pmHello,

If you make the chart, you would then have to go through and cross out anything duplicates (1x2x3, 1x3x2, 2x3x1, 3x2x1, 3x1x2, 2x1x3) because the represent the same thing. There is not easy way to get ride of the duplicates when making a chart.

addmanJuly 14, 2017 at 12:32 amits complicated

KumarJune 1, 2017 at 3:04 amGood explaination

PJMay 3, 2007 at 7:33 pmThanks alot! This was actally a better explanation then my teacher could give us =]

no-oneSeptember 21, 2016 at 10:18 amhallelujiah!

AllyFebruary 26, 2017 at 9:04 amyea, same!

anwarOctober 27, 2017 at 11:38 amConclusion

With the above discussion, it is clear that permutation and combination are different terms, which are used in mathematics, statistics, research and our day to day life. A point to remember, regarding these two concepts is that, for a given set of objects, permutation will always be higher than its combinations.

poo lordOctober 28, 2017 at 7:59 amwhat r u talking about?

bleahFebruary 28, 2017 at 4:20 pmAgreed – this article helped me understand it WAY better.

AasthaDecember 29, 2017 at 7:50 amYeah nd that too in a kinda fun manner….as if explaining a kid. Great work!

KalidMay 3, 2007 at 10:14 pmAwesome! Glad you found it useful :)

AnonymousMay 10, 2007 at 7:13 amfinally this makes sense

ConfusedMarch 29, 2017 at 3:11 pmThis still makes 0 sense

Hannah VMarch 31, 2017 at 4:56 pmI understood the permutations. But then I got confused at the combinations (I think I got confused because they used the combinations for the same example as the permutations)

If you pick 3 people from a group of 10 (Example 1 in the *A few examples* paragraph.) wouldn’t you do 10 x 9 x 8 like in the permutations paragraph, but then because unlike a permutation, a combination doesn’t care if you mix up those 3 numbers so you would have to divide 3! from 10 x 9 x 8.

10 x 9 x 8 = 720

3! = 3 x 2 x 1 = 6

720/6 = 120

That is what the author got, and I understand that.

If this helps, my teacher explained it using the word GLOSS. IF we were to find the number of ways we could mix up its letters, we would do 5! because there are 5 letters in the word. You put one of the letters (L for instance) in first spot, which leaves 4 letters left (G, O, S, S). You get another letter (say it’s S) in second spot which leaves 3 letters left (G, O, S). And so on until you have 0 letters left. So you started off with 5 letters, and went down to 0, which is why you do 5 x 4 x 3 x 2 x 1 or 5! But there are two S’s and it does not matter where you put either S. SOLGS is the same as SOLGS, right?! I swapped the S’s but you don’t notice. Now, because there are two S’s, we have to divide 5! (possible combinations) by 2! (the 2 S’s that are interchangeable)

5! = 5 x 4 x 3 x 2 x 1 = 120

2! = 2 x 1 = 2

120/2 = 60

Therefore there are 60 possible combinations to arrange the letters G, L, O, S, S.

If we use the word EXCEED, we would do 6! (because there are 6 letters) divided by 3! (3 E’s) which is:

6! = 6 x 5 x 4 x 3 x 2 x 1 = 720

3! = 3 x 2 x 1 = 6

720/6 = 120

For EXCEED there are 120 ways we could arrange those letters.

Mind you, this was only for combinations.

If we did permutations…

For the word GLOSS, it would matter where we put the S’s and they would NOT be interchangeable. Then it would just be 5!, which is 120.

For EXCEED, the E’s would not be interchangeable either, so it would be 6!, which is 720.

I hope this helped you understand Permutations and Combinations better, and if not, writing this helped me understand it better, so thank you!! :)

AlexiaMay 15, 2017 at 3:13 amthat was very helpful thankyou

shyamOctober 30, 2017 at 1:40 amPlease explain in how many way we can arrange 5men and 5 women so the 2men can not come together………

I know the answer 2880…but please explain how….

LameckNovember 24, 2017 at 11:39 pm5 women can be seated in a circular manner such that there is a vacant between them

(5-4)! since its circular you subtract 1

Now 5 men can fit in the remaining seats with no two adjacent men

4!*5!

=2880

KennyMay 22, 2007 at 4:08 pmthis is an awesome site!

SheriMay 25, 2007 at 10:31 amThanks a million! It makes sense now!

KalidMay 25, 2007 at 11:53 amThanks for the comments, glad you found it useful.

D CooperJune 5, 2007 at 6:18 amIf my chances are 1 in 13 million of winning the lottery and I buy 10 tickets, do my chances increase?

bill the math manOctober 27, 2016 at 6:49 amyes they will be 10 in 13 million kinda obvious

ShaneJanuary 25, 2017 at 3:28 pmKinda late to answer that one, plus its funny how you go on forums and find out your teachers have been using the same shit for 10 years

who knows?February 7, 2017 at 7:00 amyes

KalidJune 5, 2007 at 10:40 amHi D, when you buy multiple tickets you would add up the chances. So 10 tickets would be 1/13,000,000 + 1/13,000,000 + 1/13,000,000 … = 10 / 13,000,000

So buying multiple tickets would increase your chances for that particular lottery. If you somehow bought half of the available tickets, you’d have a 50-50 chance. And if you bought all of the tickets you’d win :).

SeanSeptember 23, 2016 at 8:46 amDid Cooper win the lottery after all?

R DavidJuly 17, 2007 at 7:03 amI’m having a stupid moment. I have a problem: how many combinations exist when one needs to select a team of 22 players from a squad of 40 players?

IS this 40!/22!(18!) = 113,380,261,800?

It seems a rediculaously large number! Please help me!

Abhishek SinghNovember 21, 2016 at 3:41 amSo what!! These are the number of ways .

KalidJuly 19, 2007 at 2:29 amHi David, yep, you got the formula right. The number of permutations (ways to order 22 people of 40) is:

40 * 39 * 38 … * 24 * 23 * 22 * 21 * 20 * 19 = 40! / 18!

[Be careful of off-by-one errors, I had a mistake at first. 40 to 19 is 22 people (just like 40 to 39 is 2 people, even though 40-39 is 1)]

And the number of ways to re-arrange 22 people = 22!

So we divide the first by the second and get

40!/18!(22!) = 113,380,261,800

113 billion does seem huge, but there’s a lot of multiplications happening. There’s 56 ways to pick 3 people from 8, which seems pretty large as well.

It’s one of those things where human beings (all of us!) aren’t great at intuitively estimating the impact of exponential growth. The birthday paradox and the effect of compound interest are other examples of this. I think it’s because we don’t encounter such mind-boggling growth or large numbers in a way we can really experience (at a certain point, millions, billions, and trillions become “a lot”, even though a trillion is a *million* times bigger than a million).

AlishaMay 13, 2016 at 6:50 pmThis helped with my question… sort of

Aaron ChaonJuly 30, 2007 at 4:42 pmGreat Stuff! You should write a book!

KalidJuly 31, 2007 at 3:15 pmThanks for the encouragement Aaron! Once I have enough posts I would love to turn it into a book :)

DheerajAugust 1, 2007 at 3:05 amit was really useful dude!!!

thnx a lot!!!

KalidAugust 2, 2007 at 2:39 amNo problem, glad it helped!

Brock LesnarDecember 17, 2016 at 9:10 amWorst planning

sidFebruary 12, 2018 at 12:36 pmBrock, bad comment

sam eismontAugust 23, 2007 at 10:44 amI play in a fantasy footfall league. I can select players for my team and they each earn points based on their performance in each weeks actual football game. I compete against other teams owners in my league and the owner with the most points each week wins. Also the points earned each week are totaled at the end of the year and the owner with the most points wins the annual point competition.

Of course there are limitations and rules to the game. Of all the players listed I may select only 22 players. Of the 22 players the team must be composed of:

3 quarterbacks (QB)(58 QBs)

6 running backs (RB)(81 RBs)

6wide receivers (WR)(125WRs)

2 tight ends (TE)(56 TEs)

3 kickers (K)(37Ks)

2 defenses (D)(32Ds)

Also each player is assigned a salary and my salary limit for the team is $60,000,000.00 for all 22 players.

I can trade for additional players each week but I’m limited to 120 trades for the year.

Here is the question?:

??? Of all the players available which ;

2QB+6RB+6WR+2TE+3K+2D whose total salary does not exceed $60,000,000 will generate the most projected points?????

The program should list the top 20 combinations in descendinding order of points.

Attached is a file that lists all players available. Of all the columns available the only ones used will be Name (player), Salary (in thousands) and PNTS ( projected points for the in todays game).

I think your program PermutCombine will do some of the work but I’

Thanks again for your interest. Sam Eismont

TomAugust 29, 2007 at 1:49 amif a train has 18 cars , and 3types of cargo must be transported, how many ways can the 3 types be transported if one type of cargo can at most occupy only ten cars per train?

BeshoSeptember 24, 2007 at 3:36 pmThanks alot!!!!

am studying n i have an exam 2mmorow !!

thnx 4 helpn me

KalidSeptember 27, 2007 at 9:23 amHi Besho, I’m glad I was abble to help!

SahilSeptember 30, 2007 at 12:12 am10 pairs of shoes are well mixed up.4 shoes are randomly picked. What is the probablity of getting at least 1 complete pair

ChrizApril 24, 2017 at 12:51 am10c1*18c2

UtkarshFebruary 21, 2018 at 2:21 amHi Sir!

I ran some calculations and found the answer to your question as

10C1*9C1*16C1*2+10C2

—————————

20C4

PrasOctober 2, 2007 at 6:03 amThis is really an excellent way of explaining the things!!!

KalidOctober 2, 2007 at 7:55 amHi Pras, thanks for the comment!

@Sahil: It’s a good question, but I want to make sure I’m not doing someone’s homework for them :).

In general, it’s easier to find the chance of “zero matches” and subtract this from 1, vs. finding the chance for 1 match.

So let’s find the chance for zero matches. Imagine picking your first shoe, A: nothing special here, you aren’t going get the match on a single shoe.

You pick shoe B: You have a 18/19 chance of getting zero matches (only A’s partner would match, of the 19).

You pick shoe C: You have a 16/18 chance of zero matches (only A and B’s partner would match, of the remaining 18).

You pick shoe D: You have a 14/17 chance of not getting any matches (only A and B and C’s partner would match, of the remaining 17).

If you multiply these chances you get the total chance for zero matches. Subtract from zero to get the chance for any match. At least I think that’s how it goes :)

ChrizApril 24, 2017 at 12:58 amNo. Of ways a pair can be picked from 10 – 10c1

No. Of ways 2shoes (any) can be picked from 18(remaining) – 18c2

No. Of ways picking atleast one pair in 4 out of 20 = 10c1*18c2

I hope it’s crct, pls correct me if I’m wrong

safaOctober 6, 2007 at 3:32 amhey

I was about to crazy solving this sum which i now find was actually so simple thanks :)

safaOctober 6, 2007 at 3:37 amI feel the answer to this is simple but i am just not able to get it..

In an examination there are three multiple choice questions and each question has 4 choices. The number of sequences in which a student can fail to get all answers correct is..

KalidOctober 6, 2007 at 7:58 amHi Safa, for that question it helps to take it one step at a time.

In a test with only 1 question, how many ways can you be wrong? 3. (Suppose the right answer is D… you could answer A, B or C).

Now how about 2 questions? Well, you have 3 ways to get the first question wrong, and another 3 ways to get the second one wrong. So the total is 3 * 3 = 9. (Let’s say the right answer is D and D. Then AA AB AB BA BB BC CA CB CC are all wrong).

Similarly, if you have 3 questions, then there is 3 * 3 * 3 = 3^3 = 27 ways to get all answers wrong. (You can write it out but will take a while: AAA AAB AAC ABA ABB ABC ACA ACB ACC… you get the idea :) )

Hope this makes sense,

-Kalid

nelNovember 30, 2016 at 5:27 pmhi!!! where can i find the solutions to the problems stated above???

Tundaghishe amwaipingaAugust 25, 2017 at 8:28 amHello

AnonymousOctober 24, 2007 at 11:38 amthanks too much

i understood it

thanks again

splitlineOctober 30, 2007 at 8:00 pmNice explanation. I was wandering if you could explan some more about COUNTING….

Thanks..

splitline..

KalidOctober 30, 2007 at 11:06 pmHi Splitline, thanks for the suggestion — I may cover counting in the subject of an upcoming article.

At a high level, to count the number of ways to do something, you multiply all the choices together. So, if you want to count how many ways to get 3 cards in poker, you’d do 52 (first option is to pick any card) times 51 (second option is any of the remaining cards) times 50 (third option is any of the leftover cards).

This is the general idea — a full article may be needed to make it more clear.

GeethaNovember 14, 2007 at 10:25 amIt was a very useful site,indeed!

KalidNovember 14, 2007 at 1:02 pmThanks Geetha!

HanyNovember 15, 2007 at 7:30 amHi

I would appreciate if you answer this:

A survey question has 6 answers, you can choose a single answer or any combination from the 6. How many possible combinations are there?

Thanks a lot

Hany

KalidNovember 15, 2007 at 11:03 amHi Hany, this question is a bit different. In this case you have 2 choices (use or don’t use the question), and you make this decision 6 times. So the number of possibilities is

2 * 2 * 2 * 2 * 2 * 2 = 2^6 = 64.

Update: A great point was made below (comment #163) — the case of ZERO questions should be removed. So you have 64-1 = 63.

HanyNovember 16, 2007 at 10:20 amThank you very much Kalid, you made my day!

What is the exact term for this type of calculation?

Now, I want to put a formula for this in Excel to automatic coding of the 64 possibilities; is there a way to do that?

Thanks a lot

Hany

AnonymousNovember 17, 2007 at 9:58 pmdidn’t help, im more confused

KalidNovember 18, 2007 at 1:26 amSorry it didn’t work for you — try to forget it if you’ve become more confused :).

This is more of a refresher for people that learned combinations and permutations but then later forgot the formula [like me]. If you’re learning this in class, try running through a few examples in your textbook.

KathleenNovember 18, 2007 at 3:38 pmPlease help! How many combinations of wins are there in 12 football games?

JonNovember 19, 2007 at 2:59 pmIs this part of a field called ‘combinatorics’ or is that something totally different?

If so, could you do another explanation in that field, I have been reading your posts and this one and the ones on e and ln are terribly interesting.

KalidNovember 19, 2007 at 3:09 pm@Kathleen: I’m not sure if I understand the question.

If counting the number of sequences [Win-win-win-lose-lose…], you can win or lose each game. You have 2 choices at each game, and 12 games, so there are 2^12 = 2048 possibilities total.

If you’re counting the number of different records (6-6, 12-0, etc.) then there are only 13: 12-0, 11-1, 10-2 … 1-11, 0-12 [it’s 13 because we’re counting down to 0, not 1].

@Jon: Glad you liked the articles! Combinatorics is about the number of ways to “count” something (from the wikipedia article), so permutations and combinations would fall under that title.

Permutations/Combinations also occur in statistics, when you try to find the likelihood of a certain event happening out of all possible events [and you need to count the number of possible events].

Given the counting questions here, I’ll add another combinatorics article to my topic list :)

JonNovember 21, 2007 at 11:57 pmThank you very much Kalid.

I started as a philosophy major, and decided to go into computer science/A.I. for my master’s where I have been discovering an unexpected love for the beauty of mathematics. And it makes me smile to see sites like this one with open forums and quick feedback for interesting topics. Thank you again, and this will definately be a site I check regularly!

KalidNovember 22, 2007 at 7:59 amHi Jon, thanks for dropping by! You’re more than welcome — there’s so much beauty in math, programming and other topics that is often buried under dense proofs. I’m glad you like the site, I want it to be an open forum for learning :)

ChrisNovember 24, 2007 at 11:31 pmGreat posts… but I have another question:

How many 4-letter combinations are there of the letters in each word? a) ONOWAY b) OSBORNE c) OUTLOOK

I’ve been fighting with this for about 3 hours now. The answers in the text are a) 11 b) 25 c)15

I can’t figure out how to manipulate the formula to account for the duplicate letters. please Heelllllp :)

KalidNovember 26, 2007 at 4:26 pmHi Chris, great question. This is a tricky one that had me thinking for a bit. Consider a) ONOWAY at first. Pretend that the “O”s are different: there’s O1 and O2.

To find regular 4-letter combinations, do

C(6,4) = 15. That assumes the “O”s are different. Because they are the same, we need to subtract duplicate items like

“O1″WAY and “O2″WAY

How many duplicates do we have? Well, we find the number of ways to have an O and some combination of the remaining letters. We need an O + 3 other letters (chosen from 4):

C(4,3) = 4

Once we subtract off the duplicates we get:

C(6,4) – C(4,3) = 15 – 4 = 11

For b), we would do

C(7,4) – C(5,3) = 35 – 10 = 25

I’ll leave c) up to you :).

muraliNovember 27, 2007 at 2:14 pmI have stumped by this one! Can anyone help, please?

Lisa lost the combination to the safe where the

secret cookie recipe is held. She sent for

Bill Becker, the most prolific safecracker

in the prison system, and offered him a

royal pardon if he succeeded in opening

the safe.

After several attempts at bypassing the

combination, Bill realized that the only

way to open the safe is to try every possible

combination by hand. The special lock

has a four-character code. Two of the

characters must be letters, and the lock is

case sensitive (with AB not the same as

ab). The other two must be digits,

anything from 0 to 9.

What is the maximum number of

combinations that Bill would have to

try before finding the correct code?

GiridharNovember 29, 2007 at 2:06 pmThankz kalid, The site is so very cool.I am glad to visit this.

KalidNovember 29, 2007 at 2:27 pmThanks Giridhar, glad you found it :)

AnonymousNovember 29, 2007 at 6:04 pm1.How many different creations can you create all together using one ice cream flavor and at least one mix-in. (there are 52 flavors and 33 mixins)

2. How many different combinations of pizzas can you make using at least one topping including crust options. ( 5 crusts, 17 toppings)

katieNovember 29, 2007 at 6:19 pmmy question is the one above! i need help.. asap thanks

Naushad AliDecember 1, 2007 at 5:52 amThanx a bunch, loved it!

KalidDecember 1, 2007 at 10:01 am@katie: This sounds a bit like a homework problem; I think I’ll have to do a follow-up on counting techniques.

@Naushad: Awesome, glad it helped :)

DennisDecember 1, 2007 at 1:31 pmI have a hw problem that seems to involve both permutation and combination. Can you make any suggestions on how to put it all together? :

100 people / 4 prizes; two of “this”, and two of “that”. How many ways to award the prizes if a person “x” wins one of “that”.

So, I see that there are 99 left in the pool, and that two of the prizes of the 3 remaining are the same, so combination is in needed and permutation. But how?

DennisDecember 1, 2007 at 1:32 pmSorry, THANKS!

KalidDecember 2, 2007 at 1:28 amHi Dennis, I’ll take a quick stab. If I understand right, there’s 100 people and two prizes (A and A) and two other prizes (B and B). I assume there’s 1 prize per person.

First, just think about giving out 4 random prizes (A B C and D). You’d just pick 4 winners from 100: P(100,4)

This is a permutation because the order matters — prize A is different from prize D.

However, this doesn’t take the duplicate prizes into account. There’s 2 ways to arrange the Bs. There’s 2 ways to arrange the As. So, we need to divide by 4 to handle these combinations (2 is simple enough no formula is needed, but technically 2 = C(2,1)… how many ways can you pick 1 item (the item to swap) from 2?).

(Note: if all prizes were the same, we’d divide by 4 * 3 * 2 * 1, instead of 4, and end up turning the permutation formula into a combination).

Hope this helps. (And hopefully I didn’t mess it up).

@Katie: I realize I should give you a hint to get started.

For the ice cream, you’re going to get quite a large number. First, you have 52 choices for ice cream.

Next, for each mix-in you can decide to leave it in our out. That is two options per mixin, for 2^33 options total. You need to subtract 1 because you can’t leave all the mixins out. So you’d have something like

52 * (2^33 – 1) which is a pretty large number.

The pizza question would be similar.

DennisDecember 2, 2007 at 4:10 pmThanks for your time Kalid (on Sunday nonetheless). You know, now with just 3 weeks to go, I can safely say that Discrete Math has presented me with more headache than Linear & DifEq combined….lol.

OK, I’m still uncertain. 1 of four is accounted for. Thus, I understand that if B,C,D were distinct then it would be as simple as P(99,3). From above, my little mind extracted P(99,3)/2 since two of the prizes are the same. Not quite a straight Permutation or Combination??? AHHHH!

DennisDecember 2, 2007 at 4:20 pmP.S. To what ends does this site address. I just found it yesterday, and I’m quite impressed. I enjoyed the explanation above and the view of previous post. I have many friends coming up behind me that I will inform of this site. And as for myself, just 1 left Probability, Stats, & Modeling.

Thanks;

Dennis

DennisDecember 2, 2007 at 5:59 pmWait a NY minute… if person “x” wins 1 of 4 prizes, and because of duplicates P(4,1)/2 = 2. Do I then get 2*P(99,3)/2 = P(99,3); again dividing by 2 for duplicates?

Thanks.

KalidDecember 4, 2007 at 2:15 amHi Dennis, thanks for the comments. Yep, this site is about any topic that has given me or others grief, though usually on math/programming/business/communication topics (as I’m most interested in those).

I think I just thought of an easier way. Suppose we pick the “winners” first and then hand out the prizes. There are C(99,3) ways to pick 3 winners from 99.

Let’s call them 1, 2 and 3. We can distribute the remaining prizes (2As and 1 B) like so:

123

_______

AAB

ABA

BAA

So, we have 3 * C(99,3) possibilities. I think :).

ShakaraDecember 4, 2007 at 6:04 pmI don’t know what it is, but this subject is not staying in my head. I just don’t get it. I honestly can’t see the difference between the two…….I’m going crazy, but I need to learn this stuff. Help!

KalidDecember 5, 2007 at 4:50 pmHi Shakara, you might have to read this explanation (and others) a couple of times. To me, I think about whether the order I pick people makes a difference. For some things (picking 1st, 2nd, 3rd) the order matters, for other things (just making a group of 3 people) the order doesn’t matter.

If the order matters, then there’s “more ways to pick” since you could have done it one of several ways.

Frank LDecember 9, 2007 at 1:39 amI am having a difficult time with this and I have a test tomorrow. I’ve read several examples but my problems confuse me. My HW asks:

How many ways can a teacher pick four students from a class of 20 to clean up after a party?

How would I do that problem?

Also, how do I compute P(6,3) and C(3,3)?

I’m so lost right now.

BenjaminDecember 9, 2007 at 8:36 pmSir,

My question is “if the probability of a company’s pen manufacturing defects were 1/10, and if 12 such pens were manufactured, what would be the probability of the following:

1.)exactly two would be defective??

2.)at least two will be defective??

3.))none will be defective??

I am not asking you to do my homework for you but i dont want to show you all the solutions i tried and take up space. just to prove i tried working on it i tackled it the following way (i am sure i am wrong).:

ans 1 => mean = 2*(1/10),variance = 4*.1, and s.d = .2 – .4 = -.2…..now how do i get p(x=2)??/?

PLS HELP!

thanks and regards,

ben

Mohammad AliDecember 13, 2007 at 4:27 amThis is a really cool website and it also addresses permutatons and combinations which is my worst topic ever. Kalid I never seem to understand this topic. no matter what. The best so far has been your small intro to this topic but even after reading your explanations whenever I try new questions on this i get stuck. Can you please give a detailed post on this topic of combinatorics. I will be very very very grateful.

Mohammad AliDecember 13, 2007 at 4:28 amI dont know why but whenever I start doing these questions its like a wall comes up in my mind….do you think i need to think more on these questions? My basics?What could be the problem?

KalidDecember 13, 2007 at 7:04 pm@Frank: Picking 4 from 20 would be a *combination* because you don’t care the order. In this case, you’d plug in k=4 and n=20 into the combination formula above n!/[(n-k)!k!]. k is the number of items you want, and n is the number of total items.

@Benjamin: This is more of a stats problem, but I’ll give some high-level points. The 3) is easiest: you need to find the chances that all pens worked well. The chance of 1 pen working is 9/10, so the chance of every one working is (9/10)^12.

For the other questions, it helps to invert. For the chance that at least 2 are defective, you can think about the chance exactly 0 or 1 pens are defective, and take the opposite probability. These can get a little tricky to compute — I’ll probably have to do a post on it.

@Mohammad: Thanks for the suggestion, it seems people would like a more detailed look at these. I’m not an expert but have found a few techniques that work for me. A lot of familiarity comes with practice — start with easier problems and work your way up. I’ll be sure to do a post on this topic in the future :).

AztralDecember 16, 2007 at 12:59 pmHere’s one i’ve been pondering since yesterday…

—-Lining up marbles —-

Let’s say you have 3 bags of marbles. Bag 1 has m different marbles, bag 2 n different marbles, bag3 l different marbles. You may

How many different ways to line up the marbles in a row of 3 (you may only use 1 marble from each bag?

DennisDecember 16, 2007 at 10:04 pmHi Kalid….It’s been a while. I just wanted to say thanks again. With a final on Friday 21st I’m a little nervous. I do plan to read through the site a few more times.

Kalid, with regards to Aztral’s post (Don’t go by me Aztral)can we say:

1) There is a total of 3 positions.

2) Choosing form bag 1, 3 choices to place m marbles i.e. (3m)

3) Choosing form bag 2, 2 choices to place n marbles i.e. (2n)

4) Choosing form bag 3, 1 choice to place l marbles i.e. (l)

Leaving us with (3m)(2n)(l)?

This sounds like something similar to what I might see………don’t know

DennisDecember 16, 2007 at 10:09 pmRecursive definitions and algorithms. Any suggestions on some links.

Thanks

KalidDecember 16, 2007 at 10:31 pm@Dennis/Aztral: Yep, you guys are on the right track. There’s a few different ways to think about problems like these, I really need to do a follow-up :)

I first forget about the order the marbles. If you have 3 bags (M, N, L), then the total choices are

M * N * L

Using real numbers: If I have 10 Maroon marbles, 5 Navy Blue, and 3 Lime, there are 10 * 5 * 3 = 150 choices.

But we didn’t talk about the order. For any 3 marbles, ABC, we can re-arrange them 3 * 2 * 1 = 6 times:

ABC

ACB

BAC

BCA

CBA

CAB

So, we have to multiply our 150 arrangements (where M was picked first) by 6, to get 900.

Similarly, you’d have 6 * M * N * L. You got the same result through a different path, which is great. The key is to recognize the impact of the permutation (ordering).

AztralDecember 18, 2007 at 7:59 amHi everyone.

Thanks for the help.

I’ve been reading up on set theory ever since, and came up with this (I also realized I didn’t state that a) the marbles are unique, b) the selected marble from bag1 always goes in the first position, marble from bag2 in the second…ie. no need to consider arrangement since they’re already arranged)

Let’s call the bags “sacks” now ;), so that sack1 is S1, sack2 is S2,….

Then basically we’re just creating a new set S=S1xS2xS3. |S| = |S1|x|S2|x|S3|

I appreciated the help :)

KalidDecember 18, 2007 at 11:57 amGreat, glad you figured it out :). Yep, that’s one way to look at it — if the arrangement is already fixed, you have S1 x S2 x S3.

JonathanDecember 23, 2007 at 6:21 pmI need help with this problem,

A drawer contains eight red, eight yellow, eight green and eight black socks. What is the probability of getting at least one pair of matching socks when five socks are randomly pulled from the drawer?

Thanks

KalidDecember 24, 2007 at 5:31 pmHi Jonathan, that’s a bit of a trick question — try doing an example where you pull out 5 random socks and see what happens :).

NeilDecember 30, 2007 at 1:40 pmGreat website!

Following up on your response to #41 above…

I am looking for a generalized formula for combinations when one has to select r items out of n items, where there can be z items that are similar in the original n items, with frequencies k1, k2, … kz.

I saw a formula on-line that says the answer is

n!/(k1! k2! … kz!) but this doesn’t take into account r.

Please help!

Also, does this class of “similar items” apply to permutations as well? If so, is there a generalized formula for permutations too, when there are z similar items in n original items, and one is taking them r at a time?

Thanks!

Neil.

TlnaJanuary 1, 2008 at 10:24 pmCan anyone give me the answer to this question.

HOW MANY 7 LETTER GROUPS CAN BE MADE FROM THE WORD”ARRANGEMENTS”

AnonymousJanuary 3, 2008 at 3:43 pmi like it a lot

bobJanuary 3, 2008 at 3:43 pmi do to

DaveJanuary 4, 2008 at 9:49 amHere is one that a number of us have been pondering for some time. Suppose I was just dealt two hearts from a standard deck of cards. What are the odds that exactly 3 of the next 5 cards dealt will also be hearts? There are 11 hearts remaining in a deck of 50 cards and I want exactly 3 of them in the next 5 cards, and the ‘set’ seems to be boolean, Heart or Not. It seems like quite a different problem from standard combinations. I’ll keep working on it and let you know if I solve it.

DaveJanuary 4, 2008 at 10:43 amNumber of possible hands matching my criteria is 11_C_3 * 39_C_2 = 165 * 741 = 122,265 possible hands with three more of the same suit. Divide that by the number of possible hands 50_C_5 = 2,118,760 and we see that I have 5.77% chance of getting exactly 3 more hearts. Additionally there are 12,870 remaining hands with 4 hearts and 462 remaining hands with all 5 hearts, so starting with two hearts in my hand I seem to have (122,265 + 12,870 + 462) / 2,118,760 = 6.4% chance of making a flush with suited hole cards.

DiegoAndresJAYJanuary 4, 2008 at 1:31 pm“I’ve always confused ‘permutation’ and ‘combination’ — which one’s which?”

I was working at a quick service restaurant (we had combo meals) when I first learned combinations/permutations in school. I found it helped me to think that when a customer ordered a combo meal, just like with combinations it didn’t matter what order they received each item — just that they were all present.

KalidJanuary 5, 2008 at 12:30 amHi Diego, thanks for the comment — that’s a nice way to visualize it.

MattJanuary 11, 2008 at 8:14 pmKalid,

As a new math teacher, I’m always looking for new ways to help my students understand tough concepts. It’s nice to find someone else that doesn’t believe in just memorizing formulas. Keep up the good work; I’ll be checking in here often!

KalidJanuary 11, 2008 at 8:25 pmHi Matt, that’s great! The world needs more teachers who focus on more than memorization :). Good luck!

TopiJanuary 16, 2008 at 9:15 pmThe posts have been extremely useful. But I cant figure out how to work this one. If you have 5 cards and one particular card must not be at either end. I know its apermutation problem but I cant figure out what to permutate.

TopiJanuary 16, 2008 at 9:26 pmI found out the answer(48) but I still cant figure how it was worked out

TopiJanuary 16, 2008 at 10:00 pmActually the answer is 72 and it worked out by per mutating the cards without the special card in all the different possible positions, which in this case is three. that gives 24 3 times which is 72.whew.

WynerJanuary 17, 2008 at 4:15 amHow many 4-letter combinations are there of the letters in c) OUTLOOK?

Would you please solve this one? Thx.

Michał LewtakJanuary 26, 2008 at 3:01 pmPeople could skip classes and learn here instead. It’s the same thing, it not better. Great job!

Michał LewtakJanuary 26, 2008 at 3:02 pmif*

KalidJanuary 26, 2008 at 4:46 pmThanks Michal, glad you liked it!

RajeshJanuary 30, 2008 at 3:47 amGreat stuff, find in great help!

KalidJanuary 30, 2008 at 7:52 pmThanks Rajesh, happy you found it useful.

navedJanuary 31, 2008 at 12:04 amits a very helpful site for maths students

KalidJanuary 31, 2008 at 12:07 amThanks naved!

javad majdFebruary 27, 2008 at 2:17 amThe customer can order pizza from 8 topping. from no topping to 8 topping. How many different kind of pizza can a customer order from no topping to 8 topping

christineMarch 3, 2008 at 7:26 pmthanks for this website. it helped so much. :D

and i have a test tomorrow on it about this and i was freaking out. now this makes me feel so much confident. thanks a billion. xD

ChristineMarch 3, 2008 at 7:32 pmHang on, I need some help. it says: Art, Becky, carl, Denise, and Ed all want to go to the concert. However, there are only 3tickets. How many ways can they choose the 3 who get to go to the concert?

i kinda know how to do it ..i just want to make sure. I wanted to make sure why it’s combination: because the order doesn’t matter right? whatever order the 5 ppl are/ the 3 tickets (since the tickets are all the same) it would be combination right?

KalidMarch 3, 2008 at 7:47 pm@Christine: Glad the site was helpful!

You got it — it’s a combination because the tickets are all the same. Giving tickets to Art, Becky & Carl is the same as giving it to Carl, Becky and Art.

Now, if the tickets were different (front row, second row, and third row) then it would be a permutation since ABC (Art in the front row) is different from CBA (Carl in the front row).

Barry GMarch 5, 2008 at 7:21 amHi,

I have the following problem: Supposing I have 4 horse races and have to select the winner in each race (I can select more than 1 horse in each race) and I have selected 1 winner from race 1, and 2 winners in each of the last 3 races, is there anyway method or nice way to list out all possible permutations between these races? I know there are 8 permutations (1 X 2 X 2 X 2 = 8) but I’m trying to figure how I can write these 8 permutations out without error and without having to go through each possible sequence mentally…I know it’s relatively straightforward for only 8 perms but if there were 16 perms (2 X 2 X 2 X 2) it will be a lot more complicated and so on. Just wondering if you can think of a gerneralised method to write these down accurately.

Race 1: A

Race 2: B X

Race 3: C Y

Race 4: D Z

I’d greatly appreciate any help on this, cheers.

KalidMarch 5, 2008 at 10:20 amHi Barry, great question! I actually just wrote about this very technique:

http://betterexplained.com/articles/how-to-understand-combinations-using-multiplication/

One method is to turn it into an equation:

a * (b + x) * (c + y) * (d + z)

And that will show you all the possibilities. Calc5 can do symbolic calculations (example here).

Barry GMarch 6, 2008 at 11:19 amKalid, thanks a million for your answer – it’s exactly what I was looking for! Keep up the good work!

KalidMarch 6, 2008 at 11:33 amHi Barry, you’re more than welcome — happy you found it useful.

migzMarch 8, 2008 at 7:21 pmCan u help me solvong this problem?

A singer practiced seven different songs. he plans to sing four of them for a television program and other three for a radio special. In how many ways he can sing them? if he does not want to repeat any of the song?

(I need it now…hope you can help me)

KevinMarch 11, 2008 at 11:48 amthanks, I had strep and missed school and this lesson so the site really helped! =^..^=

KalidMarch 11, 2008 at 12:05 pm@migz: Sorry, looks like a homework question — try asking your teacher!

@Kevin: Thanks, glad it was helpful!

venkatesha prasadMarch 22, 2008 at 2:00 amreally, this is the way one should explain the concepts. thnx a lot, lot, lot……..

KalidMarch 22, 2008 at 7:09 amYou’re welcome Venkatesha, happy you found it useful.

someone could help me please!!March 24, 2008 at 11:03 ami have question on permutation,, please if somebody could help me answer it,,, i am blank on it

this is the question

” how many ways to create a line of 10 women and 6 men, but dont make two man stand each other…??”

plsss…,, helmi, indonesian

TinaMarch 31, 2008 at 2:00 amI have 2 scenarios:

1st:

I have 99 numbers from 1-99, i have to make groups of 15 numbers, numbers can be repeated but not in same group & all numbers should not be repeated. How many such groups can be made?

2nd:

I have 90 numbers from 1-90, I have to make groups of 15 numbers, there should be atleast 1 number from 1-10, 11-20, 21-30, 31-40, 41-50….81-89

How many such groups can be made?

Ur help is highly appreciated.

DominicMarch 31, 2008 at 8:11 pmThanks but I still don’t get it…

KalidApril 1, 2008 at 12:18 pm@Tina: Sounds a bit like a homework question :).

Hi Dominic, sorry it didn’t help you — if you can be more specific about what parts were confusing I can try to make those sections more clear.

HarleyApril 9, 2008 at 6:39 amYou have a choice of 10 main dishes, 8 side dishes, and 13 desserts. How many combo’s are possible?

I’m really stuck on this. I /think/ that k is 3, but I can’t find n for the life of me.

NeemithaApril 15, 2008 at 6:08 pmthanks so much for putting this in such a simple manner. I could never make sense out of any of it until I saw ur explanation…thanks!!!

KalidApril 15, 2008 at 6:37 pmAwesome Neemitha, I’m glad it worked for you :).

Another Good article on Permutation and Combination - The TestMagic ForumsApril 23, 2008 at 12:06 pm[…] Another Good article on Permutation and Combination Easy Permutations and Combinations | BetterExplained […]

DylanApril 25, 2008 at 9:12 pmHeya,

I’m struggling to understand how combinations work with regards to binomial series and such. In post 41, using C(4,3) to determine there were 4 different ways the O could have been represent still puzzles me. Its easy to see that you could have Oxxx, xOxx, xxOx, xxxO, and so the 4 needed to be subtracted, but i still can’t make sense of the factorial approach. It comes up with coin tosses (binomial), where to choose the number of successes (n) out of ten trials is 10!/(10-n)!n!. I see that as the formula saying there are 10! possible outcomes, but really there are only 2^10? Can you make sense out of this for me please?

KalidApril 26, 2008 at 12:48 amHi Dylan, that’s a really great question. The difference between “counting”, “ordering” and “grouping” can be really subtle and it’s confused me plenty of times.

I consider “counting” to be finding the number of ways something could happen. Ten coin tosses does have 2^10 = 1024 ways of playing out: you have heads/tails each throw (2 choices) and multiply this 10 times.

“Ordering” (permuations) is finding the number of ways to pick an ordered subset from a larger set: like picking gold, silver and bronze from 10 people.

“Grouping” (combinations) finds the number of ways lets you pick an unordered subset from a larger set.

Let’s take a look at the coin toss example. One question is “how many total possibilities are there?” With 10 throws, you have 1024 possible outcomes.

But we actually want a different question: How many ways can I get exactly n winners? I don’t care about the *total* possibilities, I just want to count the number of ways to get exactly 3 successes, for example.

Let’s start off easy: how many ways can we have 0 successes (aka all tails?). There’s only one way to do that: Get all tails.

Mathematically, this is C(10,0): having 10 items and choosing none of them. In this case, 10! is the number of ways we could order the coins — we have 10 choices for the first coin in line, 9 for the next, 8 for the third, etc. It doesn’t matter if it’s coins or people: there are 10! ways to order 10 items.

If we want to get exactly 1 winner, we have C(10,1) = 10 choices. If we want exactly 2 winners, we have C(10,2) = 45 choices, and so on. In fact, our pattern looks like this (try it out online):

1 + 10 + 45 + 120 + 210 + 252 + 210 + 120 + 45 + 10 + 1 = 1024

Crazy, eh? But it should make sense — the total # of ways to have 0, 1, 2 … 10 successes should be our total number of possibilities.

Phew. So in summary, counting problems (2^10) lists out the *total* ways of doing something, while combinations/permutations let us count out specific ways of doing something. It can be confusing, so thanks for bringing it up. I think this needs to turn into a post :).

CarolynnApril 26, 2008 at 11:56 amHelp! How many ways can you get at least 7 out of 10 answers correct on a test?

KalidApril 27, 2008 at 2:27 amHi Carolynn, for this question you’d think about having 7, 8, 9 or 10 answers correct. So you’d have

C(10,7) + C(10,8) + C(10,9) + C(10,10)

DylanApril 27, 2008 at 5:13 amThanks a lot for that Kalid. It still took me all day to understand and accept the distinction, think i do now.. the factorial 10! is the number of ways you could order 10 successes. if you wanted 3 successes, and you calculated P(10,7) the number would include winning the first second and third, and winning the third second and first (it dosn’t make sense to view it as a sequence of events, although it makes it clear why you would use combinations, because its impossible to have won the third before the second). this is sorted by dividing by 3!. And thats why 10! is such a large number compared to the logical 1024. Does any of that make sense, or is my reasoning flawed?

KalidApril 27, 2008 at 9:54 amHi Dylan, that’s exactly it! You explained it in terms that made sense to you, and I think you’ve got it. Permutations are really nitpicking (first, second, third is different from third, second, first) and dividing by 3! helps get rid of these similarities.

In the case of coin flips, there’s really only 1 order, so combinations make sense. But if you just walked around and saw 10 coins on the ground (that had already been flipped), then permutations may make more sense: you could visit first, second, third or maybe second, first, third and see 3 winners each time.

Your reasoning is right on. This is a tough distinction and it has confused me plenty of times (For example, I had forgotten that all the combinations add up to 2^10).

choleApril 29, 2008 at 2:21 pmit is not what i leared!!at school.. change dis !!im just more confused then i was before!!now do something. all my friends and even my teacher said this was a bad exmple!now im sorry to be so harsh be this really sucks…Now try to think like a kid and change it..

thank you so much my dear i love you!!!

KalidApril 29, 2008 at 2:28 pmHi chole, sorry it didn’t work for you. The examples are more of a review than a start from scratch — you may want to revisit them after studying the lesson in your book :).

EnricMay 2, 2008 at 1:00 pmHi Kalid! Thank you for your awesome post!

Can you help me with probabilities in poker?. Wikipedia says probability to get three of a kind is C(13,1)*C(4,3)*C(12,2)*C(4,1)^2, but I can’t figure out why they use combinations, and in such way… Thkx

BenjaminMay 14, 2008 at 5:52 pmYESSS!! thank god there is a website like this. this is exactly EXACTLY what i wanted.

THANK YOU!!!

KalidMay 14, 2008 at 7:47 pmThanks Benjamin, I’m glad it helped you!

CarlyMay 19, 2008 at 4:58 pmThis is so awsome. I have EOG’s tommorow and I really needed a remider on this stuff, thanks!

KalidMay 19, 2008 at 10:55 pmExcellent, glad it was helpful!

Dave R.May 20, 2008 at 4:52 amSomething has confused me a little. It appears to me that you’re using the slash as the subtraction operator and dot as the multiplication operator.

Is that an American thing (I’m British), a mathematicians’ thing, or am I just confused by programming languages?

KalidMay 21, 2008 at 7:51 amHi Dave, great question. When writing text, I use star (*) for multiplication and slash (/) for division. This is the format used by most programming languages.

When making graphical equations in the articles (as above), I try to use a horizontal bar for division (-) and a dot for multiplication since that is how many people write it out (in America at least).

In the example above, 8!/5! means

8 * 7 * 6 * 5 * 4 * 3 * 2 * 1

divided by

5 * 4 * 3 * 2 * 1.

If I’ve used slash (/) instead of (-) for subtraction that would be a mistake on my part.

KBMay 27, 2008 at 10:38 amThank you so much for explaining it in such simple terms. Sure to book mark this link.

KalidMay 27, 2008 at 11:55 amYou’re welcome KB, glad it was helpful!

HariJune 2, 2008 at 9:52 amthnx a lot.i understood the theory clearly only now.again thnx a lot.

KalidJune 2, 2008 at 11:05 pmThanks Hari, glad you enjoyed it!

gerbsJune 4, 2008 at 3:42 pmur rokk so much i am 4ever in ur debt. i have a final tomorrow and i forgot my book but thnx to u i can rememeber permutations and combinations

KalidJune 5, 2008 at 11:03 amThanks gerbs, I’m happy it was useful — good luck on your test!

Sbs MatematikJune 6, 2008 at 3:58 pmThanks these are very usefull for me

gurneetJune 7, 2008 at 1:55 amhi, the explanation was great!

i just had a small doubt .

say a n letter word half “x”s and half “y”s

xxxxxx……yyyyyy……

what are the total word permutations for this word. someone told me the answer but i dont know how the formula works. could u plz plz explain the working of the formula.

taking the SATIIJune 7, 2008 at 6:11 amthanks i knew these were on the sat subject tests but i couldn’t remember how to figure them out and this brought it all back, thanks again

MaheshJune 8, 2008 at 11:24 pmIt helped. Simple explanation, but effective. The word redundancy in Combination made the things simpler.

june aceJune 10, 2008 at 3:21 amwow!!! this really is a fantastic site,thanks for the help on defining what is permutation and what is combination,i got a better idea of it now.thanks so much

KalidJune 10, 2008 at 11:07 am@sbs: Glad it was useful.

@gurneet: Interesting question. The fun part about combinations/permutations is thinking about how to set up the problems.

In this case, let’s pretend you have all x’s, i.e.

xxxxxx

Now we want to change half of them to y’s. We just need the number of ways to choose 3 items from 6, or C(6,3).

We’re doing a combination because it doesn’t matter the order we change those x’s to y’s — we just need it done. It’s like the tin can example above: we’re picking 3 people to be “special” and turn into y’s.

In general, the formula would be C(n, n/2).

@taking the SATII: Glad it was useful!

@Mahesh: Thanks!

@June: Thank you, glad it was helpful!

charbel k.June 12, 2008 at 11:39 pmhello, can you help me finding a formula in excel so i can get all number possibility from 6 number going from 1 to 42,some kind like the loto possibilities, (1,2, to , 42) and choose 6 numbers

AnonymousJune 18, 2008 at 6:50 amI ‘ve also enjoyed ur explantion but there seem to be more complications the more complicated the selection and/or ordering has to be done. I will like u to send me a detailed treatment on this stubborn topic of combinatorics. Consider this question for example and explain its solution to me:

How many 4-permutations from the set of the first 100 positive integers exist, if 3 of the integers in each permutation are consecutive integers in their usual order?

AaronicJune 18, 2008 at 6:54 amI ‘ve also enjoyed ur explantion but there seem to be more complications the more complicated the selection and/or ordering has to be done. I will like u to send me a detailed treatment on this stubborn topic of combinatorics. Consider this question for example and explain its solution to me:

How many 4-permutations from the set of the first 100 positive integers exist, if 3 of the integers in each permutation are consecutive integers in their usual order?

(Forgot to add my email!)

ankastreJune 19, 2008 at 1:06 amThank you very much Kalid

Sangramsinh TakmogeJune 22, 2008 at 4:35 amLook at permutation this way.. It is about things Per Mutation i.e. all possible ways things can take shape. :) What say?

JUNEJune 26, 2008 at 8:34 amUMMM….my friend really troubles me with this….Here is the question:

a boy wanted to buy a shirt which cost 97 pesos..

his mother and father gave him 50 pesos each….

he bought the shirt and got a change of 3 pesos…

he gave 1 peso to each of his parents….

50-1=49…if he only got a debt of 49 pesos to each of parents…..which counts it as 98 pesos….and he got the other 1 peso…..which makes it 99….

where is the other one peso(i am not sure if there is any)and why is it there? and how is it possible that there is a missing number if he bought a 97 peso shirt with his 100 peso?…..

my answer in here is that the missing peso is kept by the boy but i am not sure how to explain it……i think it is something like AB=BA……PLEASE HELP

program images webmasterJune 30, 2008 at 10:31 amThank you so much kalid…

AnonymousJuly 6, 2008 at 11:47 pmyep its rght the boy has keot 1 pesos

SidJuly 9, 2008 at 12:29 pmThis is one fantastic website

kudos to u guys

keep up the gr8 work

seviye belirleme sınavıJuly 16, 2008 at 5:33 amif he only got a debt of 49 pesos to each of parents…..which counts it as 98 pesos….and he got the other 1 peso…..which makes it 99….

anonymousJuly 16, 2008 at 8:29 amWow! Excellent explanation. I really understood this.

blackstar104July 16, 2008 at 8:30 amPermutations and combinations – never thought I would understand them.

Generate and Print Bingo / Housie tickets using Excel Sheet | Pointy Haired Dilbert - Chandoo.orgJuly 16, 2008 at 10:50 am[…] Instead we can list all the possible “5 cells with numbers and others are empty” combinations of a 9 cells region and select a random combination every time. There are essentially 9C5 i.e. 126 ways in which you can select any 5 cells out of 9 cells (without repetition of course). […]

SharmaJuly 27, 2008 at 7:16 amThanks alot for sharing it across. Soperb explanation!

Regards,

Sharma

SharmaJuly 27, 2008 at 7:17 amThanks alot for sharing! Superb explantion.

Regards,

Sharma

SharmaJuly 27, 2008 at 11:12 am1. 10 reply is not correct.

40*39*….22*21*19 gives 127439496778816000000000000000000 value.

Further we cant end above in 18 ie., 40*39*…19*18 . If we take this way it is actualy 23 people not 22 people.

Please change accordingly.

Thanks,

Sharma

RickJuly 28, 2008 at 6:26 pmHow many 6 digit combimations are ther from 17 through 47? Can you send me them? Thanks

Prat ShellJuly 30, 2008 at 5:55 pmI have to create a 3 digit number using numbers from a set having 4 numbers – 1,2,3,4.

nPr = n!/(n-r)!

=> nPr = 4!/(4-3)!

=> nPr = 24

I understand the above.

But what if numbers are allowed to be repeated, i.e., numbers like 111,222 etc are also allowed. I have physically counted and I think I can have 64 3-digit numbers using the digits 1,2,3,4. Is this permutation or combination?

And what is the formula to get the answer 64?

Help me pleeaaase?

KalidJuly 30, 2008 at 6:11 pmHi Prat,

In this case, you don’t need permutations/combinations, just multiplication. Call the digits A, B and C.

You have 4 choices for A: 1, 2, 3 or 4

You have 4 choices for B: 1, 2, 3 or 4

You have 4 choices for C: 1, 2, 3 or 4

So you have 4 * 4 * 4 = 64 possibilities in all!

Combinations/permutations work well when you have items that aren’t repeated, like people in a group. For things like numbers (where you can have as many 1’s as you please) then multiplication does the trick. Hope this helps.

KalidJuly 30, 2008 at 6:12 pm@Sharma: Whoops, thanks for the correction! I’m fixing it now.

Prat ShellJuly 30, 2008 at 6:19 pmMany Thanks for your clarification Kalid. Couldn’t be simpler than that !!

You have one more fan here, and I would be coming back time and again to see your increasing fan club. Have you thought of posting Youtube videos? :-)

KalidJuly 31, 2008 at 3:45 pm@Pret: Glad you enjoyed it! I’ve started thinking about making quick videos, maybe I’ll try that out :).

antonioAugust 9, 2008 at 10:01 amhi–

i was trying to do something with sets that have repetitions. f.ex. (a, b, c, c, d, e, f, g, g) how many ways are there to order this? i know that fex. (a, b, c) would be 3!=6. but how it would work with repetitions? any idea?

thanks

AntonioAugust 9, 2008 at 11:20 amHi,

I´ve found something that explains my question:

http://www.regentsprep.org/rEGENTS/MATH/permut/LpermRep.htm

it´s called permutations with repetitions, but i still don´t know how the formula could be explained more easily or have a proof. Some commentary on that would be welcomed…

thanks,

AshuAugust 19, 2008 at 12:32 pmhow many possible ways are there to arrange the letters of the word “PERMUTATIONS” in such a way that there will be exactly four letters between “P” & “S”

PradeepAugust 22, 2008 at 9:32 amYour answer at #31 must be corrected to 63.

The question says, ‘you can choose a single answer or any combination from the 6’, which means you choose at least one answer. The case of no answer must be considered.

MarkAugust 27, 2008 at 9:40 pmGreat stuff.

Re the pesos question: it’s a trick question and you are getting it backwards.

You start with 100 pesos, but you should subtract 2 (the amount returned to the parents) from the original amount, not add it to the shirt

ie 100 – 2 = 98

So the new starting amount = 98.

Add the shirt (97) + the boys 1 peso (1) = 98

My question is, is it still a combination when the number in the group can be any value?

eg how many groups OF ANY NUMBER can be made out of a set of five people.

I can add it up easily enough as:

1 group of 0 (total = 1)

5 groups of 1 (total = 6)

10 groups of 2 (total = 16)

10 groups of 3 (total = 26)

5 groups of 4 (total = 31)

1 group of 5 (total = 32)

…but what formula would I use to get that 32?

If I had 100’s of people it would be much harder to calculate manually.

btw – isn’t it neat how that group is symmetrical – 1,5,10,10,5,1 – does that follow for all group sizes?

CSSeptember 5, 2008 at 10:20 amOk, I still can’t figure out this problem. My professor said it was a combination,but I still don’t understand how to apply the formula to this problem.

How many different ways can you make change for a $50 using $5 bills, $10 bills, and $20 bills? I know the answer, but I don’t know how they got it.

Sari MarksSeptember 13, 2008 at 1:50 pmI learned combinations and permutations as part of Business Statistics in college. I know that, according to the formula, there are 720 possible permutations for a lock numbered “0” through “9” (ten numbers) and having 3 dials. What I don’t understand is why there aren’t 1000 permutations–just starting with “000” and ending with “999.” I just count 000, 001, 002, etc. through 999. Why doesn’t the permutations formula work out for this? This is driving me crazy–Please Help!

Sari MarksSeptember 13, 2008 at 2:03 pmNever mind–I found my answer by looking back at #156. I don’t use permutations, just multiplication. 10 choices for the first dial, x 10 choices for the second dial, x 10 for the third. 10x10x10 gives 1000 possible lock settings using 3 dials having 10 numbers on each dial. Thanks.

priyaSeptember 14, 2008 at 5:51 ami think that your blog is amazing!! i was absen for the class that my teacher did P&C and now i totally understand it!!! i really want more!!i think that you should do more advanced stuff under imp topics like calculus etc. it would help me and everyone else a lot!!

KalidSeptember 16, 2008 at 6:20 pm@Mark: Great question — you’re basically discovering the binomial theorem :). In essence, it is 2^N, where N is the number of decisions you make.

So,in a group of 5 people you’d have 2^5 = 32. With a group of 6 you’d have 2^6 = 64. The reason is that for each item, you decide “in or out?”, a decision with 2 choices. That means with 5 items, you had 2^5 or 32 possible outcomes. I plan on doing a follow up article on this point. (Also, it’s neat that you can break it down into individual groups of 0..5 and have the sum work out).

@CS: That isn’t a typical combination problem because it can change a few ways. You might have to break it down into steps: 50 = 20 + 20 + 10, which is one result that can be broken down further. I’d have to think more about how to do this “cleanly”.

@Sari: Glad you figured it out! Yes, when you can choose ANY digits, then you have 000-999, or 1000 answers.

If you must choose 3 DIFFERENT digits, you have 10 * 9 * 8 = 720, the original number. Hope this helps!

@priya: Glad you enjoyed it! Thanks for the suggestion — I’m doing a few articles on Calculus now, if you check out the latest posts on the homepage. Hope you enjoy them!

ljwSeptember 28, 2008 at 2:54 amu really did help me more than my maths teacher! rawk on dude…

KalidSeptember 28, 2008 at 11:41 am@ljw: Great, glad you enjoyed it!

MattSeptember 29, 2008 at 8:49 pmif have 3 cereals to choose from and can add any 48 different ingredients to those cereals (with a maximum of 6 ingredients), how many different cereal mixes can I make? I need to see how to do this problem?

Mahmoud HeshamOctober 8, 2008 at 11:52 pmThanx alot Kalid :)

you actually explain it better than my teacher

KalidOctober 9, 2008 at 4:49 pm@Matt: This type of question would be a good follow up article. For specific help, try the Dr. Math Forum: http://mathforum.org/dr.math/

@Mahmoud: You’re welcome!

transpaletOctober 13, 2008 at 6:32 amGreat, Thx…

online haberOctober 16, 2008 at 6:28 amHaber

Thanks so much from haberler. Haber means in Turkish is News

anisaOctober 20, 2008 at 7:56 amI have a statistical problem which is killing me… i try posting it – there would be some chance to get an answer right?? as the b’day paradox – let’s hope!! ii begging for some reasonable help…please!!

“six cups and saucers come in pairs: 2 pairs are red, 2 white and 2 blue. if the cups are placed randomly onto the saucers (one each), find the probability that no cup is upon a saucer of the same pattern”.

KalidOctober 20, 2008 at 11:46 amHi Anisa, a better place for this type of specific question is probably http://mathforum.org/dr.math/.

Thanks,

-Kalid

anisaOctober 20, 2008 at 2:09 pmThank you very much Kalid!!

have a nice day,

Alex CameronOctober 21, 2008 at 11:58 amHi D,

I need your help!

The problem asks:

Four students have to be chosen. 2 girls as the captain and vice captain and 2 boys as captain and vice-captain of the school. There are 15 eligible girls and 12 eligible boys. In how many ways can they be chosen if Sunita is sure to be the captain?

Thank you,

Alex

AyushOctober 23, 2008 at 8:30 amFinally some action on P&C, I should tell ya Kalid, some time back in high school I missed the classes on P&C and till now I’ve been perplexed with this topic, another one to mention is Sets.

But voila, this post has been amazing to read all the way. I can say the topic looks less confusing, but I have my own set of problems.

@Alex

Sunita = 1

Now total no of girls availabe for vice captain= 14

So, ways of selecting vice captain = C(14,1)

Ways of selecting boys = C(12,2)

Hope I’m correct.

ABOctober 30, 2008 at 4:17 pm@Alex:

I could be wrong, but here’s my stab at it:

For the girls, the Captain is already chosen, so only 14 girls are left out of 15. So:

– Number of ways of selecting the Girl Vice Captain = C(14,1), or 14

– Number of ways of selecting the Boy Captain and Vice Captain = P(12,2), or 132

So, total possible ways they can be chosen is 14*132 = 1848. Note, I think that for the Boys, you must use a permutation versus combination because order matters.

AnonymousNovember 1, 2008 at 10:46 amAlex, i guess you are correct. selecting a captain and a vice captain from n boys makes it a permutation problem. I am telling this after reading what kalid has said above. in fact he has given a simiar example. If it were only selecting 2 oys fron n boys, it would be combination, but here the order matters, as, (captain, vice captain) is different from (vice captain, captain).

crystalNovember 11, 2008 at 8:59 amthanx,so if i am to say number of arrasngements by taking a letter “a” at the beginning of each arrangement from a word like “alkaline”

and then what about number of arrangements taking the first and last letters as “l”…

MERCYNovember 12, 2008 at 2:41 pmIT IS AN AMAZING MATHS . I WISH TO ANSEW THE QUESTON.BECAUSE IT GIVES ME ENCOURAGEMENT OF DOING IT THANKS.

DerekNovember 16, 2008 at 12:50 amMy son has 3 green balls, 3 gold balls and 4 silver balls.

He asked me how many unique patterns can he make by lining them up in a row using all 10 balls each time.

Can anyone help please?

LauraNovember 16, 2008 at 2:05 amMy son has 3 green balls, 3 gold balls and 4 silver balls.

He asked me how many unique patterns can he make by lining them up in a row using all 10 balls each time.

Answer: 10!/(3!3!4!) = 4200

KalidNovember 16, 2008 at 6:06 pm@Derek: Good question, and @Laura: thanks for the quick reply!

One way to think about it:

If *all* the balls were different colors, there would be 10! ways to arrange them.

Since 3 are green (G1 G2 and G3), we need to divide by 3! (= 6) to factor out the redundancies:

G1 G2 G3

G1 G3 G2

G2 G1 G3

G2 G3 G1

G3 G1 G2

G3 G2 G1

These 6 situations are the same from our perspective, so we need to divide them out. Similarly, we need to remove the redundancy with gold (3!) and silver (4!). So the final answer is:

total possibilities / redundancies

10! / (3! * 3! * 4!)

which is what Laura wrote. For me, the hardest part for these problems is knowing how to “set it up” — in this case, take all the possibilities and divide out the redundancies (vs. starting from zero and trying to add up the unique situations).

AbdullaNovember 23, 2008 at 12:38 pm@ Kalid: Nice explanation,but I don’t know why I’m always confused with Permutation and Combination problems.. eventhough I memorized the formula but I can’t use the correct one.. :(

KalidNovember 23, 2008 at 2:42 pmHi Abdulla, I know what you mean — I often confused which one was which. Perhaps instead of thinking “permutation” and “combination” think about what needs to be done — do we need to find the order of things, or groups of things in any order? (The official name of this is not important).

If we need to find the order of things, we can pick the first (n choices), the second (n-1) choices, and stop once we have enough.

If we can have items in any order, then we need to divide again by the number of redundancies. If we have groups of 3, there are 3 * 2 * 1 possible redundancies.

It may help to think about what is happening under the hood, vs. the name “permutation” or “combination”.

kalyanNovember 25, 2008 at 9:04 amdistribute 100 Rs boys should get 5 Rs gals should get 50 Ps and children should get 25 Ps total 100 persons should be there total 100 Rs should be distributed how many boys, gals, children

sadieNovember 25, 2008 at 1:41 pmpermutations are mutant combinations. Think of riding in a car with 3 friends, Mary, John and Bob. So there are 4 people in the car. Anyone can drive so it doesn’t matter where they are sitting. Then Mary gets mad because she always has to sit in the back. Now you have to mutate this and figure out how many different seating arrangements there are. There are way more arrangements when you mutate a combination.

sassaDecember 7, 2008 at 11:21 pmOn stretch of road 75miles long, two trucks approach each other. truck A is traveling at 55miles per hr. and Truck B is traveling at 80miles per hour. what is the distance between the two trucks in miles one minute before their head- on collision?

santoshDecember 8, 2008 at 5:22 ammind blowing

DeepDecember 10, 2008 at 9:14 pmthanks man, helps a bit

AnonymousDecember 14, 2008 at 3:29 pmhi its v useful website thanks 4 this i want to ask ,how can i find that a question should be solved by permutations, combinations ,or by multiplication this is a question that make me cofuse bye

tabiDecember 14, 2008 at 3:29 pmhi its v useful website thanks 4 this i want to ask ,how can i find that a question should be solved by permutations, combinations ,or by multiplication this is a question that make me cofuse bye

Aaron WillhiteJanuary 6, 2009 at 6:13 pmI am having trouble. Would you tell me how to figure the problems: 14C0 (combination), and: In order to conduct an experiment, 3 subjects are randomly selected from a group of 15 subjects. How many different groups of 3 subjects are possible? (and is it a combination or permutation?)

KalidJanuary 6, 2009 at 7:53 pmHi Aaron, in this case think about whether the order matters (permutation) or doesn’t matter.

If you picked the people for the study backwards, would it change anything?

Snow ZhaoJanuary 8, 2009 at 4:22 pmstill don’t know about it…

I don’t understand clearly..

melvinJanuary 9, 2009 at 3:28 amWonderful explanations and insights;Thanks Khalid and all ;

Got a problem, I’d like some lead ; I’m trying to develop an algorithm ; I have { 0 1 2 3 4 5 6 7 8 9 } AND want to determine how many PINs of 4 digits can I generate from that; Case in point without repetition first , then with repetition

ctlrJanuary 9, 2009 at 2:44 pm@ melvin

case 1 (without repetition): 10 x 9 x 8 x 7 = 5040

“10 choices for first number” times “9 choices for second number” etc.

case 2 (with repetition): 10^4 = 10,000

“10 choices for first number” times “10 choices for 2nd number”, etc.

AnnJanuary 19, 2009 at 6:12 amWow, this helped me a lot! thanks!

Vibhanshu BhardwajJanuary 19, 2009 at 10:29 pmProblem:- let assume there is one suitecase having 3 digit lock system and the user forgot that password. now the question is how many combination can be made for this?

thanks in advance

sandyFebruary 1, 2009 at 7:30 amat last i knew it, this was clearer than that of what our prof explained..

wHeew!

thanks a Lot!

NouriFebruary 3, 2009 at 8:28 pmAwesome explanation! Wish my teacher did it this well. I love your humor in it too! Hope to see more of my work and lessons on here! :D

KalidFebruary 3, 2009 at 9:25 pm@Ann: You’re welcome!

@Vibhanshu: You don’t need combinations for this type of question. Imagine starting with the lowest combination and counting up — how many would you have?

@sandy: Glad it helped.

@Nouri: Thanks! Yeah, I find writing about math gets boring really fast unless you find ways to spice it up :).

AnonymousFebruary 6, 2009 at 10:45 amamazing how you explain the question

AnonymousFebruary 6, 2009 at 10:51 amim from sulaimanyah a part of iraq city idont think that you ever heard about it!!!!!!!!!!!!!!! but i welly like to say that your explination is so usuful……….

How many 4-letter combinations are there in the word, OUTLOOK?..thnxFebruary 9, 2009 at 7:43 amThanks..great forum.

SandyFebruary 10, 2009 at 10:04 amMy daughter had this problem on a test..there are 4 pork , 5 beef, 3 chicken and 3 noodles dishes. How many combinations of food can you order? She’s never done combintions before. Help

MathtutFebruary 12, 2009 at 9:03 pmwhen you have groups of combinations you simply multiply the number in each group together

Lets begin with only the first two. 4 pork dishes and 5 beef dishes. There would be 20 combinations between those two(4 *5)=20. Now with those 20 combinations in mind,lets say you add 3 chicken dishes. Now how many combinations can you have. (20 * 3)=60, because for each combo of pork and beef you have 3 different chickens choices. So now we are up to 60 combos of pork , beef and chicken dishes. Now lets add 3 noodle dishes and yep you guessed it, there would be (60 * 3) combos for all 4 dishes, for a total of 180 combinations. You could make up the same type of problem with clothes. Take some outfits of your daughters, say pants, belts, shoes..etc. Lay them out for her and figure how many of each you have. multiply out how many combo’s there should be and go through those combos physically(keep it small so it doesnt take very long perhaps 3 of one item, 2 of another, lastly just one God Bless Mathtut

SandyFebruary 13, 2009 at 3:32 pmThanks so much!!

KalidFebruary 13, 2009 at 4:23 pm@Mathtut: Thanks for helping out Sandy!

coryFebruary 15, 2009 at 7:13 pmstupid

CourtneyAllisonFebruary 18, 2009 at 2:59 pmwhen you’re doing something, permutations, or combinations, and order DOES NOT matter, it becomes a combination, right?

CourtneyAllisonFebruary 18, 2009 at 3:02 pmcomment 210. when you’re looking at a combination of a word outlook, it would be… Permutation (7,7) because there’s 7 letters. And then, when you have the double letters, such as the O, you divide the answer of P (7,7) by 3! (factorial because then you would have extra letters in there not necessary.) What is left over, from my understanding, and i may be wrong, is 4 letters. Your answer should come up by doing that.

chizenFebruary 19, 2009 at 10:09 amcan someone help me with this~ onegai~

A shipment of 12 tv sets contains 3 defective sets. In how many ways can a hotel purchase 5 of these sets and receive at least 2 of the defective sets?

cadyFebruary 24, 2009 at 4:53 pmthanks a lot, this is way simpler than the explanations from my book *chuckles*

Regards,

Cady

cadyFebruary 24, 2009 at 5:20 pm*Shortcuts for Combinations and Permutations*

Math itself is complicated enough with the raining numbers and overwhelming equations. Therefore, creating a shortcut is a real handy.

Shortcut for simple permutations:

P(10,3)

Instead of doing all the subtract, divide, multiply and stuff, you can simply do this:

10*9*8

instead of going aaaaaaaaaaaalll the way to 1, simply stop with the first 3 numbers.

Note: The reason we stop at 3 because our “r” is 3. Therefore, if our “r” is 4, then we will stop at 10*9*8*7

To conclude this, P(10,3) is 720.

Let’s go to Combinations:

C(12,5)

First, find the Permutation of (12,5)

1. 12*11*10*9*8 = 95040

Note: Notice that we stop at 8, because our “r” is 5, so we stop at the first 5 numbers.

Then our Permutation is 95040.

Next, let’s find the Combination:

2. Divide the permutaion to 5!

95040/5! = 792

If you do all this in Scientific Calculator, this will be a breeze for you.

Well, see if it works! (^_^)

Doing all those formulas is way too much for me, but remember that if the teacher requires you to show your solution, then you may not be able to use this shortcut.

You can only use this for time wise.

~Nemo Omnia Scire Potest~

Regards,

Cady

EkumMarch 2, 2009 at 6:18 amHelp me please I’m confused.

if N N then?

I mean

suppose n=4 and k=10

then it goes like this

P (4,10) = 4!/ 6! = answer in fractions like 0.233 something

I’m confused-

does it go like we should take the bigger number as N and smaller as K?

please help and explain it : both for permutation and combination :(

thanks in advance.

BiniyamMarch 2, 2009 at 5:05 pmMy god you are a genius.. do you know how many time I read my probability book to understand permutation and combination concept.

KalidMarch 2, 2009 at 7:35 pm@Ekum: For combinations and permutations, the first number (N) is how many total items you have, and the second number (K) is how many items you want to pick. So you wouldn’t have a situation like P(4,10) which means you have 4 items and are picking 10 :).

@Biniyam: Glad you enjoyed it!

LaylaMarch 4, 2009 at 4:44 pmA multiple-choice exam consists of 14 questions, each of which has 4 possible answers. How many different ways can all 14 questions on the exam be answered? I’ve tried dividing 14!/(14-4)!, 14*4 using the fundamental counting principle and I’m clueless as to what to do o_O. Please help

LaylaMarch 4, 2009 at 4:54 pmoh yea and another ? In how many ways can 5 seniors, 3 juniors, 4 sophomores, and 3 freshmen be seated in a row if a senior must be seated at each end? Assume that the members of each class are distinct. O_O

ChetanMarch 6, 2009 at 4:19 pm@Layla, the answer to your first question would be 4^14.

Your 2nd question doesn’t state the number of seats in the front row. Assuming that all 15 can be seated in one row, having 2 seniors on either side, it would be 13!*C(5,2). Hope that’s right.

ChetanMarch 6, 2009 at 4:36 pmI’ve a type of problem that’s been troubling me for quite sometime now. I’ve been studying to take up GMAT soon, and every time the ‘ball’ problem comes up, I seem to get lost.

Ex: A box contains 2 white balls, 3 black balls and 4 red balls. The number of ways in which three balls can be drawn from the box so that at least one of the balls is black is?

a)74 b)64 c)94 d)20 e)24

Kalid, please help.

AnonymousMarch 8, 2009 at 2:00 pmWell this problem left some bits on info behind. However I think I know where it’s getting at. If they say the first or second ball is defiantly black then it’s a given statement. So I’m going to assume that the first ball is black. Here’s what you do then:

Since there are 3 black balls put 3 in the first slot of your permutation.

3 * _ * _

There are a total of 3 slots because your asking for the ways that 3 balls can be drawn. Then we subtract 1 from the total of balls (9 balls total) because one black ball was already taken. so it’ll be 3 * 8 * 7 = 168. 168 ways you can select 3 balls from a box.

ChetanMarch 8, 2009 at 6:44 pmThanks ‘Anonymous’. But the problem is complete and the answer is 64. I don’t know how we get there. That’s exactly where I’ve been caught up too. And the thing is, most of these ‘ball’ problems that I’ve tried solving follow a similar pattern. Which is why I began searching for sites/forums online to try and find help.

SreshthaMarch 11, 2009 at 4:10 amYes the answer is 64.This is how we get it..

no of ways =(3C1)(2C2)+(3C1)(4C2)+(3C1)(2C1)(4C1)+(3C2)(2C1)+(3C2)(4C1)+(3C3)=64.

CathleenMarch 11, 2009 at 5:17 pmI have a question for which I know the answer, but I can’t figure out how to make the formula make sense:

How many different ways can all the letters of LEASES be arranged if the identical letter are NOT distinguishable?

The answer is supposedly 180, and the formula used was 6!/2!*2!

I guess my question is, why is this the right formula? Is this a standard combination, and if not what makes it different? All I can figure is that one of the twos comes from the two letters that are repeated, but I don’t know how or why. Can anyone help clear this up for me?

CTMarch 12, 2009 at 4:29 amYou are providing an exceptionally well done service to anyone trying to really understand mathematics, Thank You!

UnshuMarch 12, 2009 at 4:59 pmThis a great website! It explains everything so well. I rathe do this than go to school!

hiteshgarg7March 13, 2009 at 4:53 amExcellent.

JMarch 13, 2009 at 6:20 pmI don’t get why it is 8*7*6

JMarch 13, 2009 at 6:59 pmI don’t get why they are multiplied at all

KalidMarch 13, 2009 at 7:33 pm@Cathleen: 6! is the number of possibilities assuming each letter is unique (i.e. the first E is different from the second E).

But since this is not the case, we need to remove the impact of the E’s and the S’s. There are 2! ways to arrange 2 items (it’s a bit silly to write 2! since it’s the same as 2), so we divide once by 2! to remove the redundant Es, and again divide by 2! to remove the redundant S’s.

It might be easier with a smaller word like FOOD: how many ways can you write out the letters? It would be 4! (all the letters) divided by 2! because of the O’s.

@CT: Thank you!

@J: We multiply to show all the possibilities. If you have 4 options for breakfast (eggs, cereal, toast, fruit), and 3 options for lunch (sandwich, pasta, soup), how many possible breakfast-and-lunch combinations are there?

4 * 3 = 12. Try writing them all out to see that this works.

We multiply 8 * 7 * 6 because those are the number of options for gold, silver, and bronze in the example. Hope this helps.

jMarch 13, 2009 at 8:21 pmMan I must be stupid, as soon as it goes 3 dimensional I lose the image of it. Thanks for the feedback however, it did help me think of this properly, juts can’t visualize the 3D

jMarch 13, 2009 at 8:27 pmwhoaa no, there I have it. You can actually picture it as a cube. I was lost there for awhile. I really appreciate your help!

jMarch 13, 2009 at 8:31 pmsorry i meant: cube isn’t the proper word because it isn’t square.

thanks again

Easy Permutations and Combinations « Math BlogMarch 14, 2009 at 7:22 pm[…] Easy Permutations and Combinations Here is highly visual and useful page on permutations and combinations. […]

3 Unit Assessment Tuesday 17 March « MrB’s BlogMarch 16, 2009 at 12:37 am[…] http://betterexplained.com/articles/easy-permutations-and-combinations/ […]

Miguel AngelMarch 21, 2009 at 7:33 pmTHANK YOU SO MUCH!

Now I feel that I got it! This is much much betterexplained than it is in my book and from the teacher… :)

KalidMarch 22, 2009 at 12:04 am@Miguel: You’re welcome! Really glad it helped :)

iMarch 24, 2009 at 7:25 pmman this site is freakin awsome

KalidMarch 24, 2009 at 10:36 pm@i: Glad you liked it!

JordanMarch 27, 2009 at 3:06 pmHello, I have a problem that is really stumping me.

It’s how many paths can be traced out in a 6×4 grid. I start in the lower left corner and and can only move right or up until I get to the upper right corner. I see that I can only make a total of 10 moves altogether to get there, so I have to use up all 4 up moves and all 6 right moves. I know the answer is 6 choose 4 but I cannot picture why this is true! I was wondering if you have some insight into this problem??

Thanks tons for any type of help

KalidMarch 27, 2009 at 3:21 pm@Jordan: That’s a great question. Here’s how I’d think about it:

* Assume your 6 right moves are “fixed”. That is, you are going to take them anyway. The only thing you can control is when you make your up moves. Do you use them all up on the first spot? (All the way up, all the way right). Or do you use them as you go along (right, up, right, up, right, up…)

6 choose 4 means “I have 6 options and pick 4 of them (I’m not allowed to repeat)”. That means “I have 6 positions I could move up, and I pick 4”. So that’s the number of paths assuming you can’t move “up” twice on the same position.

JordanMarch 27, 2009 at 4:15 pmThanks so much for the quick response. I see what you are saying by how many ways can I choose 4 from the fixed 6, that was very helpful. But I have a hard time seeing this to be the right answer. If I am only allowed to move once on each of the four that I choose from 6 I could see it working, but since I could move all 4 on any of the 6 it seems like there should be more paths than 15? I just wish I could get a good visual on this.

Thanks again I really appreciated your response.

JordanMarch 27, 2009 at 9:36 pmIs that what the answer is supposed to be 15? I would have thought 210. I am really lost on this one.

KalidMarch 28, 2009 at 10:36 amHi Jordan, no problem — this is an interesting puzzle. I thought about it more and think I have an explanation.

Think of your choices for moving as up (u) and right (r). We need to have 6 r’s and 4 u’s… and the question becomes “How many ways can we re-arrange them?”

So, how many ways can we rearrange:

rrrrrruuuu

There are 10 items so 10! [aka P(10,10)] ways to put them, assuming order matters. But each r is the same as the others, and each u is the same as the others. So we need to divide out cases where the r’s and u’s are in the same places but re-arranged.

There are 6! ways to re-arrange the r’s, and 4! ways to rearrange the u’s. So we get

10!/(6! 4!)

which is 210 as you mention. 15 is only if we cannot move up twice in the same column. This is an interesting question, I may do an article on it :).

KalidMarch 28, 2009 at 10:41 amAlso, it’s good that you have skepticism about the 15 answer. Even drawing it out on paper, you can find more than 15 paths, so some assumption about the answer must be there.

JordanMarch 28, 2009 at 2:32 pmHello Kalid, I really cannot thank you enough for taking the time out to help me out with this.

I actually arrived at the 10!/6!x4! last night that is why I had written the answer 210. But I also thought of expanding (x+y)^10 and the term with (x^6)(y^4) has a coefficient of 210. Looking at the problem as (xor y)And(x or y)And… This modeled the problem since I had two option each time, up or over to the right (X + Y) x = up y= right.

Let me know what you think.

Jordan

JordanMarch 28, 2009 at 2:49 pmsorry that should have been x= right y= up

AnonymousMarch 29, 2009 at 10:00 amYou opened my eyes. Finally i can see the difference between Permutations and combinations.

God Bless.

KalidMarch 29, 2009 at 2:43 pm@Jordan: You’re more than welcome, this was a fun problem to think about.

I really, really like that (x+y)^10 way of looking at it! It’s a great way to turn the general AND/OR problem into an equation. It goes to show there are many ways of looking at the same problem.

@Anon: You’re welcome!

JimMarch 30, 2009 at 8:43 pmHow do you determine the number of combinations for setting up different teams when you have 16 members and you want to create 4 teams of 4 with different teams members? as an example one round is abcd efgh ijkl mnop, another is aeij bfjn cgko dhlp and another is afkp bglm chin dejo are there more and is there s formula for this type of combination determination? Thank you, Jim

lindApril 2, 2009 at 10:22 amhello, i have a question that was throw at me during an interview and i kind of fluffed it up.

There are 100 people in a table tennis (singles) competition. How many matches would have to be played in order to find out the winner?

Would appreciate any help. Thank you.

LolaApril 2, 2009 at 5:55 pmI was wondering how to set up this problem:

If you have six books and one bookend that has to go on one of the two ends, how many ways can the books and bookends be arranged?

I think I have to use 6! in the equation somewhere, but if you could help me set it up, I would really appreciate the help.

JordanApril 2, 2009 at 5:59 pmIf you have 8 quarters in a bag and 6 dimes, what is the probability that you will pull out two dimes?

Do you set it up like this?

6C1/14C1 + 6C1/14C1

KurtApril 17, 2009 at 10:17 amI have to build a unique key for a software system it will use an alpha character for the first two places and 4 digits following. Exactly how many permutations can be made using this system?

AnonymousApril 26, 2009 at 4:42 pmyou really helped me understand this much better than any of the other sites ive vistited. i am greatful for your patience and your work here. thanks!

Jorge (GEORGE)April 26, 2009 at 4:57 pmwhy is it so hard for other sites to explain this stuff as clearly as you do? anyway’s all i have to say, is that you succeeded where most failed. you took the time in figuring out the best way you could explained these details to the general pulic, and it works for me. It makes me so happy to be able to understand and so eager to want to understand, when i can’t. thanks for your simplification.

PS. im bookmarking this site!!

reedMay 5, 2009 at 6:34 pmthanks!! i have a test tomorow and this made sense! ”)

KalidMay 5, 2009 at 6:49 pm@Anonymous: You’re welcome!

@Jorge: Thank you! I really strive to make things as clear as possible (there are so many things that are needless complicated), so it means a lot that it’s working for you.

@Reed: Awesome — good luck! :)

Better Explained « Xavier Seton’s BlogMay 7, 2009 at 12:36 am[…] Statistics: Combinations & permutations, Birthday Paradox, Bayes’ Theorem, […]

AnonymousMay 7, 2009 at 3:31 pmhey thanks that really really help alot, my teacher is nuts he couldn’t explain this topic and he plans a exam about this :S but this will grant me a 100 :)

StevenMay 9, 2009 at 8:59 pmThanks, that helped a lot . Much simpler than Wikipedia.

why but ayways thanksMay 12, 2009 at 4:14 pmthis is cool som what understood helping me for the end of grade test!! :=) :-}

lauraMay 14, 2009 at 12:46 pmIm having problems figuring this out..:

Every street in Canada has a postal code mad of 3 letters and 3 #’s that alternate. How many diff post codes can be created if the 1st letter must be P, and the 1st # must be 4???

txxx

SohamMay 18, 2009 at 8:35 pmWould you want to go into greater detail ?

Especially combinations ? There are so many more formulas than the ones stated.

But thanks anyway.

KalidMay 18, 2009 at 10:29 pm@Soham: You’re welcome — I’m planning on doing more follow-up articles on more advanced types of counting. Thanks dropping in.

claudia lidwinaMay 21, 2009 at 4:45 amhua!! it’s so confusing.. >.

gurudeepMay 26, 2009 at 8:33 amIt’s really very helpful.

this article justifies the betterexplained.com

best.

Other advanced topics are expected.

Thanx!

AnonymousMay 28, 2009 at 10:23 amThis was really helped clarify all my concerns about combinations and permutations. thank you so much. my teacher did a very shoddy job.

KalidMay 28, 2009 at 12:28 pm@Gurudeep, Anonymous: Thanks!

Ashley DudzinskiMay 31, 2009 at 9:33 pmHi, thanks so much for this information. It helped me alot. I’m doing the seventh grade finals this week and I wasn’t exactly sure what the difference was between the two, and which problems I would know the differnce between. Say, if they didn’t tell you that the order does or doesn’t metter on a test question, how would I know if it was a permutation or combination?

KalidJune 1, 2009 at 12:15 am@Ashley: You’re welcome! Great question — if they don’t tell you if order matters, think about “doing” the choices backwards to forwards.

If you’re picking 3 people for a group (Alice, Bob and Charles) it doesn’t matter if you call out “Alice, Bob and Charles” or “Charles, Bob and Alice”.

But if you’re picking 3 people for 1st, 2nd, and 3rd place, it matters if you say “Alice, Bob and Charles” (Alice wins) vs “Charles, Bob, and Alice” (Bob wins).

So, part of the trick is seeing if there’s any difference when you mix up the order you pick people. Good luck!

KatieJune 1, 2009 at 6:11 pmThanks, I have a test tomorrow and this really cleared it up

KalidJune 1, 2009 at 6:40 pm@Katie: Great, good luck!

chanhankiangJune 6, 2009 at 2:10 amGreetings,

I am a year 1 high school student studying in Singapore. Recently being selected to undergo some tough maths training to prepare me for a competition. And the first chapter is permutation and combination. Came across some problems and need serious guidance and explanation. Any genius or trained professionals here that could guide and help me solve this handful of questions i would be more than glad. This will take some time. If you have answers to any of the problems Email it to [email protected]

1.The license plates of car in geometria are composed of three alphabets and fours numbers. There are 2 alphabets at the front and one at the back. The four numbers are sanwhiched in between. If the first of the numbers cannot be 0 and none of the numbers are to be registered, how many different kind of liscence plates are there?

my ans is (9x9x8x7)+(26x25x24). is it correct? and what is the differences if they change the question to how many distinct liscense plates are there instead of different kinds of liscense plates? and how to do?

2.How many 5 digit numbers can be formed using the digits 0,1,2,3,4 and 5 which are divisble by 3, without repeating the digits.

I know how to do the question if the question is without the divisble by 3 part. But with it, i am lost and blank. do we take the total outcomes – the outcomes indivisble by 3? If so, how do we find the outcomes indivisble by 3? and another question by myself. What if repeating of digits is allowed? What is the answer? I think it will increase by many times cause it would be 4x5x5x5x5 for the total number of outcomes which are divisble and indivisible by 3.

3. In how many ways can 7 boys and 6 girls be seated on a straight table if no two girls are to sit together.

I try thinking on this line: hope you can understand what i doing ( _ represents each boy)

1_2 _3 _4 _5 _6 _7 _8 (so that are 8 slots for the girls) and what do i do next? 8C6? or?

4. If the letters of the words CHASM are rearranged to form 5 letters words such that none of the word repeat and the results arranged in ascending order as in a dictionary what is the rank of the word CHASM?

my initial reaction… huh? Oh my god teacher i am 17 only!! please dont do this to me… my brain cells are dieing… could you guys dont try solving by patterns, cause usually it will take quite sometime to find it but mail it to me if you have any. THANKS

ChristyJune 18, 2009 at 7:34 pmwhoah… some of the questions people asked were so hard…. i just started this year, and yeah… the posting was really good! my problem was that i couldnt tell when order was impotant or not, for an icecream scoop problem, would the order be important? i dont think so…. unless you were picky… strawberry, chocolate, and vanilla would be the same thing as chocolate, strawberry, and vanilla, but what if you cared about what you ate first? do you think in a test or word problem they would tell you? or do you think they would expect you to know? probably combunation… Thank you Kalid!.. i was looking for good explainations and this is the best site i found =)

ChristyJune 18, 2009 at 7:41 pmoh, and couldnt you use a calculator for #10 and a lot of the other ones? unless you did… you know on the texas instrument? theres a permutation, combination, and factorial button. you probably knew that…

sorry if i spelled things wrong…

KalidJune 18, 2009 at 10:28 pm@Christy: Thanks, really glad it helped! Yep, sometimes you have to guess whether order matters or not. For chocolate vanilla strawberry, it doesn’t matter if you are getting them all at once.

But, let’s say someone asks how many different ice cream cones you can make with those 3 flavors (top scoop, middle scoop, bottom scoop). In that case order matters, so it’s a permutation!

2009 Sbs SonuçlariJune 21, 2009 at 3:19 amThanks admin

henry gulaniJune 24, 2009 at 10:17 amthis is very crazy.i have learned alot from this.i look forward to learning so many interesting ideas regarding mathematics.next time try to give me a problem to work on.good bye.

AnonymousJune 26, 2009 at 10:22 amkhjhjhkj

Sbs SonuçlarıJune 29, 2009 at 12:49 pmThanks admin..

Sbs sonuçları 2009

Günlüğüm

VigneshJuly 10, 2009 at 9:30 amI really understood permutations and combnations from the examples you cited. Now can you please tell me how to do this sum which I do not understand

If Pm stands for mPm, then prove that 1+1.P1+2.P2+3.P3+…… n.Pn=(n+1)!

P.S. note that the numbers that appear after after P (i.e., 1,2,3) are actually in subscript.

CuriousJuly 15, 2009 at 1:25 amIt is such a wonderful website. Can i ask if i buy a 6 digit number (for example 123456) and i know the permutation is 720 but if i want to have all the 720 permutation to be listed out, how to go about it?

KalidJuly 15, 2009 at 1:45 pm@Curious: Great question — I don’t think there’s an easy way aside from using a computer program to list them all out for you.

I found this:

http://home.att.net/~srschmitt/script_permutations.html

You can enter “6” for the count (6 choices) and “6” for the select (you are picking 6 items from those choices).

RodnexJuly 17, 2009 at 7:13 amwow I really loev the explaination thank you a thousand times .A frienc of mine actually did explain it to me but it did not make much sense,but your explaination is grat thanks once again……….

KalidJuly 17, 2009 at 12:50 pm@Rodnex: Thanks, really glad it helped!

shahzad Ahmed ShaikhJuly 18, 2009 at 10:53 pmwonderful method sir u gave to me i will never forget it

JasonJuly 28, 2009 at 4:27 pmMy boss just suggested that everyone in our department job shadow each other for a day… Well, there are 35 of us and while I’m job shadowing the guy down the hall, he can’t really be job shadowing me. Does this mean that order does matter? The boss seems to think this would only take 35 days, which seemed bogus to me, which is how I ended up here. I’m not exactly sure I got the math right on how many days this would actually take. Any help, Kalid?

KalidJuly 28, 2009 at 8:08 pm@Jason: Really interesting question! Depending on what your boss meant, it could take a different number of days:

1 day: Everyone (34 people) job-shadows your friend down the hall :)

2 days: Everyone pairs up and job shadows someone. The next day they switch roles (since you have an odd number of people you have a small corner case there).

35 days: Everyone has to job shadow everyone else (so you have 34 days of job shadowing, 1 day of being shadowed). This assumes 34 people can job shadow one other person.

35 * 34 is the total number of job shadow-shadowee combinations (you shadow him, he shadows you). You can only “perform” one pair per day (35/2 pairs per days) so my first guess is there is 35 * 34 / (35/2) = 34 * 2 = 68 days minimum to have every pair completed.

JasonJuly 29, 2009 at 7:49 amThanks for the explanation, Kalid! I’ll totally give you the credit for coming up with the 68-day solution at our meeting tomorrow…

KalidJuly 29, 2009 at 9:35 am@Jason: You’re welcome! I’m going to figure out if this makes sense: Let’s say there’s 6 people in the group (A B C D). How many days?

Day 1: AB, CD

Day 2: BA, DC

Day 3: AC, BD

Day 4: CA, DB

Day 5: AD, BC

Day 6: DA, CB

which is 12 (4*3) possibilities for shadower-shadowee, and we can “process” (4/2) = 2 pairs per day, for 6 days.

SumanyuJuly 29, 2009 at 12:43 pm“This raises and interesting point — we’ve got some redundancies here.”

an instead of and; //small typo!

KalidAugust 3, 2009 at 6:01 pm@Sumanyu: Thanks, I just fixed it!

tsepoAugust 12, 2009 at 8:02 amYou are better than my lecture thanks a zillion, do you have some extra explanations and examples

TakosAugust 12, 2009 at 2:36 pmHey Allid,

I don’t understand why you have 8 people but you only divide by the factorial of 3?

AnonymousAugust 12, 2009 at 8:42 pmThanks. The explanation above was very helpful.

ShekarAugust 22, 2009 at 9:19 amI came across a really challenging puzzle…would be obliged if you could throw light:

A person goes to shop with 2 bags 1 red and 1 green. He buys 10 radishes and 6 carrots. While returning he distributes them into the 2 bags such that no bag is empty. In how many ways he can do it?

shakirSeptember 1, 2009 at 7:10 amawesome

nishantSeptember 16, 2009 at 1:12 amvery helpful

sayaSeptember 17, 2009 at 10:09 pmwell said…

pls keep contributing…

Mark DameSeptember 22, 2009 at 12:32 pmHow many combinations are their of the following format:

nnnxnnn

n=letter of the alphabet (you can use a letter more than once, example: llk4bbb)

x=numerical digit from 0-9

sarika guptaSeptember 22, 2009 at 9:16 pmi didnt understand d fact dat how does order matters when v have to list down d name of d 3 desserts

red chillySeptember 24, 2009 at 6:26 ambeautifully xplained!..i wish u were my teacher..u make things so interesting..may allah bless u..thanx..take care..

well wisherSeptember 26, 2009 at 6:12 pmThnx bro !!!!!

i hv test tomorrow n this helped me a lot…..nw i guess i cn understand wt i rote in my notebook…hehe

siddharthSeptember 30, 2009 at 12:42 amwell i just happened to check out permutation after a long time …. the freat thing abt this author is tat he explains every post really great dude awesome n i wish if ther was a site where almost all mathematical related things are ttught… if u kno of auch a site plz 4ward it to my mail…. thanx bud…

good work

KalidOctober 2, 2009 at 3:38 pm@siddharth: Thanks!

Lida fx15 biber hapı ikibindokuz seo yarışmasıOctober 4, 2009 at 2:48 pmbeautifully xplained!..i

vivOctober 5, 2009 at 7:34 amthanks a lot! you made it seem so simple.

KalidOctober 5, 2009 at 6:02 pm@viv: Thanks!

AliOctober 7, 2009 at 11:48 amreally it is very nice helps me

thanks very match

need (ppt)

AndreaOctober 14, 2009 at 8:43 pmThis is amazing thank you soooooo much!! Stats test tomorrow ahhh! way to make it stick…nothing else has! greatly appreciated! I’d give you the gold medal fo sho!

KalidOctober 14, 2009 at 9:26 pm@Ali: Thanks, glad it helped.

@Andrea: You’re welcome! Really glad it was useful!

LidaOctober 19, 2009 at 10:23 pmvery nice thnks

KalidOctober 20, 2009 at 8:26 am@Lida: You’re welcome!

BushraOctober 21, 2009 at 3:58 amWow….this is an amazing way to explain what I consider to be the toughest part of mathematics…Probability and the related stuff…..!! Since we deal with things that are mostly conceptual it really is tough to grasp!

Would love it galore if you could write a post on probablity as well! :)

regards,

Bushra

Thank you for your great contribution to my learning!

KalidOctober 22, 2009 at 12:22 am@Bushra: Thank you for the kind words! You’re more than welcome, and I’m planning on doing some more about probability in the future :).

yaprak dökümü son bölüm izleOctober 26, 2009 at 3:31 pmthanks so much…

SajiNovember 1, 2009 at 6:40 pmHere’s a 3D visualization I did in processing for 3 choices from 7. I hope someone finds this useful.

/*

Permutations and combinations.

Taking 3 (visualized as dimensions) from 7 (length of side).

by Saji, ([email protected]).

License: Public domain. Attribution requested.

KEY:

Permutation: Arrangement (order matters).

Repeating: All (little cubies).

Non-repeating: All except the reds. (ie transparent + blues)

Combination: Selection (order doesn’t matter).

Repeating: Cubies with white outline.

Non-repeating: The blues.

*/

import processing.opengl.*;

float csize = 20, from = 7, startPos = -(csize * from/2),

spaceW = csize * from, bxspRat = 1.0/3.0;

ArrayList saw = new ArrayList();

void setup(){

size(300,300,P3D);

strokeWeight(2);

noStroke();

}

void draw(){

float[] see = null;

background(0);

translate(width/2, height/2);

ambientLight(60,60,60);

directionalLight(255,255,255, 0,1,0);

pointLight(100,100,100, width*(3.0/4.0),height*(3.0/4.0),400);

rotateY(radians(mouseX));

rotateX(radians(mouseY));

translate(startPos, startPos, startPos);

for(int x = 0; x < spaceW; x += csize){

for(int y = 0; y < spaceW; y += csize){

for(int z = 0; z < spaceW; z += csize){

see = new float[] {x, y, z};

pushMatrix();

translate(x,y,z);

pushStyle();

if(!seen(see)){

saw.add(see);

stroke(255);

fill(0,0,255);

}else{

fill(255,70);

}

if(x == y || y == z || z == x){

fill(255,0,0);

}

box(csize-(csize * bxspRat));

popStyle();

popMatrix();

}

}

}

saw = new ArrayList();

}

boolean seen(float[] asee){

int l = saw.size();

float[] asaw;

asee = sort(asee);

boolean ret = false;

for(int i = 0; i < l; i++){

asaw = sort((float[])saw.get(i));

if(asee[0] == asaw[0] && asee[1] == asaw[1] && asee[2] == asaw[2]){

ret = true;

break;

}

}

return ret;

}

ryanNovember 3, 2009 at 1:58 amfantastic!!!!

easy to understand

thanks a lot

kristieNovember 3, 2009 at 8:52 pmThank you to the creator of this site and the people who help answer these questions because they really did help me to better understand how to answer my questions that I have. My teacher just does not know how to explain in the way you guys do and that is a shame because it makes me not appreciate math as much as I could if it were just explained in a simpler way. Thanks!

aapiNovember 5, 2009 at 9:15 amhey the explamation is very gud and very interesting yaar……….thank u

KyleNovember 7, 2009 at 7:31 pmAmazing explanation.

Helped me a lot, so thank you =)

KalidNovember 8, 2009 at 10:39 pm@Kyle: Awesome, glad it helped!

Mosiane MoshoshelaNovember 12, 2009 at 2:22 amthis is very nice,benefitted very much from it

ChaitanyaNovember 18, 2009 at 4:39 amKalid is a lifesaver

jerseyNovember 20, 2009 at 9:14 pmthanx a lot dude…..u rock!!!…

KalidNovember 22, 2009 at 8:52 pm@jersey: You’re welcome!

AnonymousNovember 26, 2009 at 2:21 pmthnx alot dude u helped me out

RonNovember 28, 2009 at 2:16 amHi. Working on some stats for my A-Levels.

learned about combinations and permutations – got most of it figured. but then……

– 2 big questions:

1. Coin is tossed 10 times. How many different sequences are possible?

2. How many ways can the letters of CONSTANTINOPLE be arranged so that no vowels are next to each other??

IQinfiNITNovember 28, 2009 at 5:34 amtop drawer idea pal… wonder if this can b spread to kids at school level across d globe…it will, i am damn sure, break d jinx of maths being a dreaded and a tough subject…

keep up d gud work…

KalidDecember 1, 2009 at 1:22 am@IQinfiNIT: Glad you liked it! I hope teachers can use this to help students learn the concepts better :).

Dorothy GaleNovember 30, 2009 at 11:04 amI have been “faking it” on perms & coms for almost 10 years, first as an SAT prep instructor and later as a middle school math teacher. I bet I have gone through 20 textbooks trying to understand.

Finally, thanks to you, I get it!!!!! You are amazing and I really appreciate it.

KalidDecember 1, 2009 at 1:14 am@Dorothy: Wow, I’m so happy this explanation worked for you! I know what you mean about “faking it”, I felt like I memorized the formulas for so many years before they finally clicked intuitively… I’m really glad it helped you!

heDecember 2, 2009 at 12:20 aman ice cream store offers 20 different topping combinations, each composed of two different items. Determine how many actual toppings there are

i don’t get it T_T

the answer is 5 but how do u get it?

cubesDecember 4, 2009 at 9:01 amgoing back to comment 41 with the user that posted how many different ways can you arrange the letters from onoway into groups of 4. i do not understand your algorithm because you say you are only interested in groups that have an O and 3 other letters in them, but words like OONW will be double counted. could you please expand on this?

i did it by counting how many distinct ways you can form groups with 2 O’s ((4*3 * 2*1)/(2!*2!)=6), 1O (4) and no Os (1)

VarunDecember 6, 2009 at 10:27 amhey i have a problem

of a lot of 10 items 2 are defective.Fid the number of different samples of 4 containing (a)1 defective (b)2 defectives ???

i would really appreciate it if u answer the question,thnx

celesteDecember 17, 2009 at 6:44 pmhow do i figure or set this problem up?

A family has 5 kids, what is the chance they have 2 girls and 3 boys. Assume boy/girl is 50/50,

KalidDecember 19, 2009 at 2:06 am@celeste: First, I’d figure out the total number of possibilities: 2^5 [that is, 2 choices (boy or girl) times 2 choices (boy or girl)… and so on, 5 times].

Next, figure out how many ways there are to get your *desired* possibility: how many ways can you pick 2 girls from 5 kids? You’ll see this is the same as picking 3 boys from 5 kids also (neat how that works out :)). Finally, you can divide it out: the number you just got, divided by 2^5 is your probability.

RonaldDecember 23, 2009 at 12:53 amHi, I’m not a math guru and could really use your help. If you have 3 sets of all 26 letters of the alphabet, how many possible 3 letter combinations could you make where repetition is allowed?

GlennDecember 23, 2009 at 5:05 pmHi Kalid,

I’ve got an actuarial exam in a few weeks and was wondering if you could help me with a problem.

Suppose a purse contains the following coins: 3 Nickels, 1 Dime, and 2 Quarters. What is the probability of getting at least 35 cents if you select three coins?

Initially I thought the answer was 4/6: 4 possible ways to get at least 35 cents out of 6 possible coin combos:

4 combos > 35 Cents: 2N,1Q; 1N, 1D, 1Q; 1N, 2Q; 1D 2Q

6 Possible: 3N; 2N, 1D, + the 4 above.

This is apparently not the correct way to solve this. Any tips you would have would be appreciated.

Cheers, and thanks for the great site!

Glenn

JasonJanuary 6, 2010 at 1:11 pmI am very impressed with your blog!

I am trying to determine the total number of possible ways that a process can occur; it needs to occur in a specified order of A-B-C-D-E. There is only 1A, there are 20Bs, there are 23Cs, there are 8Ds and there are 40Es. Can you tell me how many ways this process can actually happen?

Thanks

Jason

Celester SayconJanuary 17, 2010 at 1:40 ami have also the same problem on how to determine how many ways and the correct answer, its one of my major subject…the permutations and combination…i would be glad if you could really help me…i’m just still studying why is it happening…how to do it…if you could explain or an example which is easier

Chris vd MerweJanuary 17, 2010 at 10:05 amPLEASE HELP ME WIN THE LOTTERY!!!!! I need to know! I have to pick any five numbers from 1 to 45. how many possible combinations are there and if nt too many what are they or how do I determine what they are . The jackpot is 40 milion and one ticket cost 3.50. PLEASE SOMEBODY IN THE UNIVERSE HELP ME!

lidaJanuary 18, 2010 at 2:11 amHelped me a lot, so thank you =)

kabinJanuary 18, 2010 at 2:27 amthnx alot dude u helped me out

Ray LaudenslagerJanuary 18, 2010 at 7:14 amGreat way to explain Permutations. I am developing a project for my 9th grade Honors Algebra II class and this will help.

Mr. BakosiJanuary 19, 2010 at 7:48 pmThanks that was straight to the point

VanJanuary 20, 2010 at 7:26 amI’m happy this blog has made a lead on what I want to happen…

But I still have the problem as the example shown is not the same as my problem..

Here is my problem..

How many possible letter combination(i’m not sure of what term to use) will A to Z have?

Below is a possible letter combination list:

A B

A C

B A

B C

C A

C B

A B C

A C B

B A C

B C A

C A B

C B A

an so on….

I’m not sure on how I will solve this.. Please help me

AlexJanuary 22, 2010 at 2:09 pmThank you so much. The teacher explains it in such a different way that is unclear, but this is clear to me, and now when i take my test i will pass it thanks to you!

ShoshoJanuary 23, 2010 at 5:51 amHey may you please answer that question? :)

* A room has 6 doors; In how many ways can a man enter the room through one door and come out through different door.

I hope i get the answer:D

Thanks for this easy explanation it made things much more easier to me. !!!!!

opyJanuary 24, 2010 at 2:25 pmomg! dis site was rlly heplful!

thx a lot!

but i got dis problem:

Th number of triangles and qudrilaterals on P non- collinear points are equal, find P and the numbe of triangles…

i’ll be glad if u help me wit it!!!

BiancaJanuary 27, 2010 at 3:47 ami loooooooooooooooooove you!!!

JenJanuary 28, 2010 at 6:53 amThank you so much!! I was doing some last minute review and I cannot explain to you how helpful this was. Im also glad you dont approach math in a boring way, you attempt to make it fun and seem easy. THANK YOU!!

KalidJanuary 30, 2010 at 4:15 pm@Jen: You’re welcome — so happy it worked for you!

zayFebruary 1, 2010 at 7:26 amYou’re waaaay better than my Math teacher! :)

thank you so so much! :P

KalidFebruary 5, 2010 at 2:58 am@zay: You’re welcome, glad it helped!

Kevin!February 1, 2010 at 6:47 pmThere are 7 posters, but only space on the wall for three. How many ways can the posters be placed on the wall?

Can someone please help me??

DrewFebruary 2, 2010 at 9:04 amHi Kalid. Kudos for taking the time to help – there’s certainly a lot of questions :)

I’m not sure where my question fits in, but here goes.

I have 5 x $1 notes. How many combinations of the following can I buy:

2 A pples = $1

1 O range = $1

1 B anana = $2

I don’t want to double up, that is:

$1 $1 $1 $1 $1

AA AA AA AA O

is the same as:

$1 $1 $1 $1 $1

O AA AA AA AA

I’m guessing that is a combination, while accepting all (including the above example) is the permutation?

SaqlainFebruary 6, 2010 at 4:27 pmSuppose you have to paint a room, and you want different color on each wall. If the available paint colors are 16 then in how many ways you can paint the room. Please also state that whether it will be combination or permutation. I will be really greatful to you for the help. Thanks. Saqi

cynthiaFebruary 7, 2010 at 10:24 amsolve the equations for the different bicycle models that can be made daily with the desire techniques graphing, substitution, elimination, and matrix also explain how to check your solution. Model A takes 2 hours to assemble and Model b takes 3 hours to assemble Model a cost $25 to make per bike and Model b cost $30 to make per bike and you have a total of 34 hours and $350 available per day for these two models, how many of each model can be made in a day? Please help me .

DrewFebruary 9, 2010 at 10:10 am@ Saqlain

I *think* it depends on a bit more information.

It would be a permutation if, for example, each wall could be painted the same colours but in different order.

16!

________

(16-4)! = 43,680

And a combination if once 4 paints were used, you couldn’t use the same paints on different walls.

16!

______________

[(16-4)!] x 4! = 1820

But don’t trust my math! I’m still having trouble getting my head around it. It’s knowing how to convert the english words into the correct formula that’s doing my head in.

ddFebruary 11, 2010 at 7:40 pmthere are 15 technicians and 11 chemists working in a research laboratory. In how many ways could they form a 5-membersafety committee if the committee must have exactly one technician?

OzziFebruary 12, 2010 at 6:47 amIve got 49 items each numbered accordingly… If I had to arrange them into 6 different items in 1 box, how many different combinations of 6 items would I get…?

JoshCunrtFebruary 18, 2010 at 8:43 pmThanks. This explanation really help me catch up on my maths class.

Jaky AstikFebruary 19, 2010 at 6:29 pmThank you so much for the tutorial. I am happy I came here :) Thanks again.

LidaFebruary 21, 2010 at 4:44 amLida Zayiflama Kapsulu Daha Ayintili Bilgi

AnonymousFebruary 21, 2010 at 6:03 pmI understand it now! Thanks so much (:

NemayyyyFebruary 21, 2010 at 9:11 pmI still have NO idea…

ellieeeFebruary 22, 2010 at 3:29 pmTHANK YOUUUUU

this helped me sososo much!

KalidFebruary 22, 2010 at 6:25 pm@ellieee: Awesome, glad it helped!

HakemFebruary 23, 2010 at 4:42 amPretty nice..

KarinaFebruary 23, 2010 at 3:01 pmThanks so much! This was invaluable. One question: what about permutations in which the order isn’t undefined? For instance if there were letters A, B, C, D, and E, how many ways can they be arranged so that C is neither first nor last? I can’t seem to wrap my head around it :/

LidaFebruary 24, 2010 at 3:02 amGood on your Lida

I♥AngeloMarch 5, 2010 at 4:43 pmtnx for the big help!

:)

RebeccaMarch 5, 2010 at 10:55 pmHow many five digit positive integers that are divisible by 3 can be formed using the digits 0, 1, 2, 3, 4 and 5, without any of the digits getting repeated in a number?

Please help

RebeccaMarch 5, 2010 at 11:02 pmI kind of understood you take two cases viz. one where you dont consider 0 bcoz a number doesnt start with it…

There is the divisilibity by 3 rule where any number is divisible by 3 if and only if the sume of the digits are divisible…

case 2: says we dont use 3 in the arrangement…what i dont understand is why should we not use 3???

Im confused…please help

jessa marludeMarch 7, 2010 at 12:28 amthanks for the explanation, this is better than what our teacher had explained to us…

KalidMarch 8, 2010 at 11:14 pm@jessa: Awesome, glad it helped.

KarenMarch 9, 2010 at 10:39 pmI need help please!

You are dealt a hand of 5 cards from a standard deck of playing cards. Find the probability of being dealt a hand consisting of:

a) four-of-a-kind

b) a full house, which consists of 1 three-of-a-kind and 1 two-of-a-kind.

c) three-of-a-kind. (The other two cards are different from each other.)

d) two clubs and one of each other three suits.

I only got as far as a) 13*48/ 52C5 = 624/2,598,960 = .0002.

Now I know I’m supposed to use the formula nCr for the 13*48 but I don’t know how. I am getting answers and I can’t understand why….please help.

LoganMarch 10, 2010 at 9:56 amThanks for the article man. You made it very simple and cleared up alot of things for me. Also, thanks for not making this the boring surface level textbook type post. I was entertained and learned this for a big test coming up.

naomiMarch 10, 2010 at 5:14 pm5 apples are to be picked from a barrel containing 20 red, 15 green, 15 yellow apples. How many ways can you pick at least 3 red apples?

AND

How many ways can you pick exactly 3 red apples?

hayleyMarch 11, 2010 at 3:43 pmI dont know how to do this at all.:( can you help me out more? this is so confusing.

hannahMarch 12, 2010 at 4:34 amit was really useful!!!!!!!!!!!!!!

thanks a lot!!!!!!!:D

Muhabbet BüyüsüMarch 15, 2010 at 12:33 amit was really useful!!!!!!!!!!!!!!

thanks a lot!!!!!!!:D

taraMarch 15, 2010 at 6:01 amwho created this?

karenMarch 15, 2010 at 12:42 pmthanx!!! that really helpd me!!! it was a beter way of explainin than my teacher!!!!!!!!

briandMarch 21, 2010 at 7:13 pmdo you know any websites where I can answer sample questions in statistics (which focuses on permutation)?

AnonymousMarch 24, 2010 at 9:40 pmim preparing for gmat and this really really helped. thank u :)

billMarch 25, 2010 at 12:20 pmso I’ve been making word problems for my 7 year old, but instead of going through all the combos the drawn out way, I’d like to apply a formula. But the word problems I’ve been creating have variables, which complicate it a bit. What would be a good equation for something like this.

4 teachers are taking 12 kids on a field trip. Each teacher must watch at least 2 kids, but cannot handle more than 4. How many different teacher-kid groups can you make?

Where I got stuck trying to apply the formula because each teacher could be paired with 2, 3 or 4 or the 12 kids.

Any help would be greatly appreciated. Thanks in advance!

rose yuMarch 28, 2010 at 6:15 pmi just wanted to say ,

yourcool this made me lol (:

and it helped

anaMarch 30, 2010 at 7:33 amIt would be really helpful if you could help me out explaining a question..

Isaam has 11 different CDs, of which 6 are pop music, 3 are jazz and 2 are classical.

How many different arrangements of all 11 CDs on a shelt are there if the jazz are all next to each other?

Anthea TayMarch 31, 2010 at 12:36 amDear Sir/Madam, I have a question regarding combination, appreciate that you could resolve it for me. Many thanks

Question : Grace has 16 jellybeans in her pocket. She has 8 red ones, 4 green ones, and 4 blue ones. What is the minimum number of jellybeans she must take out of her pocket to ensure that she has one of each color?

ganeshApril 3, 2010 at 2:09 amThere are say 8 teams in a legue and they play each other twice. How many maches will there be? I know it is 56 but how to compute?

Hitesh ParsawalaApril 7, 2010 at 6:47 amI have a Problem as below.

There are 8 Teams overall in a Tournamen.

Each Team has to play 14 Games in all.

i.e One Team has to play against another Team Twice in a Tournament.

Hence, how many games will be played in the whole Tournament.

Please solve this issue.

Regards

Hitesh Parsawala

ChrisApril 7, 2010 at 9:48 pmI was very happy to find your site, Kalid. I do company training courses, but recently started helping my son with his year 11 – 12 maths, so need to brush up quite a bit. I will work through some of the posts on your site and give them a try. Can you check my working here?

Based on posts 40, 41, 210, 217 — How many 4-letter combinations can be made from the word OUTLOOK?

As you explained in your post #41, if the O’s are all different then we have C(7,4) combinations = 35 TOTAL COMBINATIONS

If we consider the O’s to be the same, then we get:

4-letter combinations from OUTLOOK

= TOTAL COMBINATIONS – SINGLE LETTER REPEATS – DOUBLE LETTER REPEATS

= C(7,4) – 2 x C(4,3) – 2 x C(4,2)

= 35 – 8 – 12

= 15

Alternative working (not using combination calculations)

OUTLOOK

3 O’s can go with U ,T , L, K (4 ways)

2 O’s can go with UT, UL, UK, TL, TK, LK (6 ways)

1 O can go with TLK, ULK, UTK, TLK (4 ways)

No O’s gives UTLK (1 way)

TOTAL = 15

AnonymousApril 13, 2010 at 3:06 pmBAH!

What the heck, I ask?

I understand permutations just fine; the combinations confuse me terribly. I’m normally a faster learner, too. :|

Sorry. This really hasn’t helped. >< Bleh. Time for more researching, I suppose.

SeferrApril 17, 2010 at 8:38 amTeskler..

:)April 17, 2010 at 12:56 pmthis was a great help thanks

Online Dizi izleApril 18, 2010 at 4:09 amthaknss

AnonymousApril 19, 2010 at 2:46 pmim having problems with homework. i have to figure out which questions are permutations and which ones are combinatios, then work them out. i know how to work them out. i just dont know how to figure out which ones are which.

HELP!!!!!!!!!

AntonyApril 21, 2010 at 1:29 pmHi – this site has been a great help – I actually did stats many years ago and it all feels a bit double dutch to me these days.

I now have a problem I cannot solve.

We have 8 people going away for 4 days. We are dining on 2 tables. I want to do a seating plan that ensures that everyone sits with everyone else at least once.

If I consider that there is one table, I think there would be 70 possible combinations C(8,4) = 8!/(8-4)!4! = 40320/576 = 70

However this doesn’t really help as I cannot figure out how to establish which of these combinations achieves the goal of getting everyone seated together at least once.

Can anyone help?

Thanks

tinaApril 21, 2010 at 4:56 pmhi,

so i have 8 different statistics books, 6 different geometry ones, and 3 different trig. if i must select one book of each type, how many different ways can this be done?

VikasApril 22, 2010 at 3:55 pmWOW!!! you explained this way better than my teacher!

AnonymousApril 23, 2010 at 3:36 amwow,it helps a lot.

MoniApril 24, 2010 at 10:19 amWow my math teacher tries to explain these permutations and combinations, but he misses the simplicity of it. You make it seem so easy! thanks a million

SunnyApril 24, 2010 at 1:31 pmYou are the best.

This helped me way more than my prof.

KalidApril 26, 2010 at 8:26 pm@Sunny: Glad it helped!

lauraApril 25, 2010 at 11:19 amSo the term “different combinations” may be an oxymoron, or at least potentially confusing?

KalidApril 26, 2010 at 8:24 pm@laura: Yep, it can be confusing because we’re not used to being so specific with our language (combination vs. permutation). “Different combinations” in the math sense means AB and BC when choosing 3 items, but some of us informally may consider AB and BA to be “different”.

Here’s some more on counting…April 27, 2010 at 7:38 am[…] Differences between Permutations & Combinations II Pre-Calculus […]

SarahApril 28, 2010 at 2:12 pmwow the permutation and combination helped a lot!!! tom’s my njask for math-wish me good luck!!!

AnonymousApril 28, 2010 at 5:55 pmTHANKS SO MUCh

satish narodeApril 28, 2010 at 8:51 pmThanks..

It was very usefull..

Mysterious PersonApril 30, 2010 at 3:40 amI don’t understand that at all

MohamedMay 4, 2010 at 1:14 amThanks alot, it was very useful for me

AnonymousMay 4, 2010 at 9:12 pmreally helpful………..

KimiMay 7, 2010 at 9:49 amI am a math teacher and this was awesome! Thanks so much you are fantastic!

KalidMay 8, 2010 at 10:27 pm@Kimi: Thanks, glad you enjoyed it!

MARTINMay 10, 2010 at 8:39 amif ten matches and three possible combinations to each match i.e. draw, team 1 wins, team 2 wins how can i list every possible combination for the ten matches overall. would love an answer. thanks.

MagiMay 10, 2010 at 3:28 pmIt helped especially with my test tomorrow!

elenaMay 11, 2010 at 6:59 pmthe following question has been driving me nuts for the last 3 weeks and even though my teacher has explain it to me several time i still don’t get it:

tom goes to a stationery shop to purchase a set of watercolors and a paintbrush. There are 3 different brands of watercolors each available in 2 different size packs. there are 2 brands of paintbrushes each available in 8 different sizes. How many ways Tom can choose a pack of watercolors and paintbrush?

PLEASE HELP!!!!!!

nkMay 15, 2010 at 11:09 amQuestion: If there are tags names 1 through 40 in a box and at any given time two tags are removed, what is the probability of the total adding up to 28 from any of the attempts

AnnaMay 16, 2010 at 8:21 amThank you this made it a whole lot easier!

KalidMay 18, 2010 at 9:56 pm@Anna: You’re welcome!

JasonMay 18, 2010 at 2:28 pmWow, this really helped me understand this material!

I’m going to ace my test tomorrow now!

KalidMay 18, 2010 at 9:35 pm@Jason: Awesome, glad it helped!

sucharithaMay 20, 2010 at 7:05 amthnq soo mch..this was very useful..:)

KalidMay 20, 2010 at 8:44 pm@sucharitha: Thanks!

AnonymousMay 24, 2010 at 2:22 amgr8 tutorial it was easy and fun

CluberMay 24, 2010 at 5:40 pmVery nice tutorial, well explained. Thanks.

KalidMay 24, 2010 at 5:50 pm@Cluber: Thanks!

RinSonMay 26, 2010 at 12:57 amThanks a Lot :)

This really helped a lot…

KalidMay 31, 2010 at 10:45 pm@RinSon: You’re welcome!

NekoMay 31, 2010 at 7:46 amThanks, dude..

E-mail me if you can so if I ever need your help..!

Barbara LewisJune 13, 2010 at 9:04 pmI do have a problem that somebody could help me with. We are having a contest at work in which the winner gets to go to Florida. In order to do so, we must sell a total of 120 % in sales for this month. We have to sell so many new activations, so many air cards and so many accessories in order to be considered. This contest is between 5 people. Can you tell me how many combinations we can use to reach our goal?

GabaybayJune 14, 2010 at 4:52 pmI flipping love this explanation! We’ve been working on this for weeks in math class and it gets me sooo frustrated. I’ve got a huge exam tomorrow and I know I’m rocking it now! I’m glad there’s people out there who know their stuff! :))) Thanks a tonn!

KalidJune 14, 2010 at 8:43 pm@Gabaybay: Awesome, glad it was useful! Good luck with your exam.

TrevorJune 15, 2010 at 7:06 pmHello, Kalid, I am a student in highschool trying to comprehend the Birthday problem that invloves a combination– I believe. I have been trying (and perhap over-thinking)as of how during your

Explanation of Counting Pairs,

23 people make 253 pairs, where

23*22/2= 253 pairs

I was just wondering if you could please explain this for me. :]

wondering if you could please

TrevorJune 15, 2010 at 7:09 pmPlease excuse the grammar, but the help would be greatly appreciated

I’ll buy your books for my discrete math course (it’s REALLY watered down from the college course of course though) [=

HariJune 16, 2010 at 7:26 amHi Kalid, Am really confused. Could you help out of this?

How many 4-letter combinations are there of the letters in OUTLOOK?

oliverJune 19, 2010 at 12:07 amits superb

KalidJuly 2, 2010 at 10:42 pm@oliver: Glad you liked it.

Bahar BangashJune 19, 2010 at 8:43 pmnice one lecture..i have got alot of from this lecture…

keep it up

KalidJuly 2, 2010 at 10:44 pm@Bahar: Thanks!

MaryJune 21, 2010 at 5:18 amHI Kalid,

I am preparing for GMAT. And unexpextedly I found this site once and from then on am regularly visiting this site. I should definitely thank you as I had learnt a lot from this site. Thanks a lot Kalid :)

KalidJuly 2, 2010 at 10:40 pm@Mary: You’re more than welcome, good luck with the GMAT!

saydurJune 21, 2010 at 10:48 pmhere needs more example

juliusJune 25, 2010 at 6:47 pmIn how many ways can 4 boys and 3 girls be seated in a row of 7 chairs if the boys and girls are to be seated alternately…

Resimli Pasta TarifleriJune 27, 2010 at 2:14 amf 7 chairs if the boys and girls are to be seated alternately…

ahMmJuly 3, 2010 at 5:59 amhow many ways can arranged in word PAALAM?/

yurieJuly 6, 2010 at 2:34 ami can’t figure out the difference between permutation and combination.

Combination: Picking a team of 3 people from a group of 10. C(10,3) = 10!/(7! * 3!) = 10 * 9 * 8 / (3 * 2 * 1) = 120.

Permutation: Picking a President, VP and Waterboy from a group of 10. P(10,3) = 10!/7! = 10 * 9 * 8 = 720.

if you didn’t mention the 2nd example as permutation i’ll assume it as combination too.

tahiyaJuly 16, 2010 at 6:29 amthis is very helpful. most places just explain which formulas to use not why we are using them. thanks a lot

HemaJuly 27, 2010 at 6:38 amcan anybody tell me how to solve the below mentioned problem

how many words can be formed from the word PROBLEMS by fixing P in the first position and S in the last position?

ShukAugust 2, 2010 at 3:06 amHey Kalid, Am really confused. Could you help out of this? A.S.A.P

E Q U A T I O N

(PERMUTATION)

the diagram shows 8 cars of different letters.

a)find the number of the arrangements in which the vowels are always together?

enriqueAugust 5, 2010 at 9:22 amIt took me a short while to see that your letters correspond to the first letter of each person’s name. You could make this easier to see if you either colored the first letter differently in your first graphic. Alternatively, instead of numbering the list of names, you could list them with a, b, c, etc.

Thanks for a clear explanation!

KalidAugust 16, 2010 at 3:50 pm@enrique: Ah, thanks for the feedback — great suggestion.

Guillermo BautistaAugust 5, 2010 at 4:37 pmThis is so cool. I have written an article similar to this. Readers may want to check it out:

http://math4allages.wordpress.com/2009/12/08/introduction-to-permutation/

http://math4allages.wordpress.com/2010/01/02/intro-to-combinations/

Arvind Kumar(A.I.T.S)August 9, 2010 at 10:03 pmThis is the best practical explaination….to grasp the concept in more easier way…

thanks dude..it will help us a lot from competetive exam point of view..

hemanthAugust 17, 2010 at 3:21 amHi,

I’ve a question. Please help me.

There are five letters and four post boxes. In how many ways can these letter be posted ?

Thank you,

AnonymousAugust 21, 2010 at 12:00 amhi! thanks a lot for the explanation. now i understand the lesson.

MetulaAugust 27, 2010 at 7:45 amHi Hemanth,

I believe there are 4^5 ways to post letters. Since each letter can be posted in any of the 4 post boxes, so for each letter there would be 4 ways and if you combine the ways for all these letters it turns out to be 4*4*4*4*4. That’s what I think, anyway Khalid is here to solve this if I am wrong.

AlAugust 31, 2010 at 2:29 amHi Kalid,

Great website you’ve got here, as an engineer my maths is pretty good, but statistics is one area of maths I’ve never studied. Your explanations are better than most I’ve come across on mathematics websites and textbooks.

I have a question that I wonder if there is an easy answer to:

‘How many permutations of 4 cards (value 1 to 11) sum to a value of 21?’

I can do this the long way, but I’m looking for a method or formula to speed up the process.

For example, for 2 cards there are 2 permutations that total 21:

10 , 11

11 , 10

for 3 cards there are x permutations that total 21:

1 , 9 , 11

1 , 10, 10

1, 11, 9

2, 8, 11

2, 9, 10

etc.

By the time I consider 4 cards however, this method is too long-winded and cumbersome to be of any use.

Do you know of a quicker way to arrive at the answer than simply listing all of the permutations and counting them?

Thanks.

KalidSeptember 1, 2010 at 11:28 am@Al: Great question. I’d have to give this more thought, but one approach might be to break it into subproblems.

You know that getting 2 cards to 21 means having (10,11). Now, to get 3 cards to 21, instead of starting from scratch, why not start from the first two (10,11)? Choose one of the cards to “break apart”, for example 10 => 4,6. So you get ((4,6), 11). I think one approach would be to figure out how to recursively break apart the pieces (You can again break 6 into (1,5)).

Off the top of my head, that’s my first instinct for approaching it :).

haripriaSeptember 7, 2010 at 6:09 amyou’ve achieved making me understand

KalidSeptember 13, 2010 at 2:23 pm@haripria: Glad it helped.

RejinaSeptember 17, 2010 at 2:05 amAwesome!!!!!!

LuannSeptember 17, 2010 at 4:06 pmI have the numbers 1, 3, 5, 7 as scores on the rings of a dart board. I have 4 darts. How would I find the number of scores possible? I know total scores 4 to the 4th power possible=256 possible scores. However some scores repeat, such as getting 1117=10 pts being the same as getting 7111=10pt. How do I eliminate those repeating scores?

nikleshSeptember 21, 2010 at 10:39 pmhey kalid , this is niklesh(nik)

can u tell me easy way for following problem,

i need the sum of 4 digit nos greater thn 4000 from following nos 0,2,4,7,9??

BillSeptember 22, 2010 at 7:52 amHi Kalid. I was wondering if you wanted to bet on 12 NFL games, how many different possibilities are there? I’m trying to see if you can bet every possibility, and the odds might allow you to win no matter what. Thanks for the help!

KalidOctober 8, 2010 at 10:21 am@Bill: If one team plays 12 games, it can either win or lose each one (ties are possible but rare, so we’ll ignore them for simplicity). That means you have a choice of 2 outcomes (win or lose) and make this decision 12 times. So, the answer is 2 * 2 * 2 … 12 times, or 2^12 = 4096 possible outcomes.

AnonymousSeptember 24, 2010 at 6:19 amcan u please answer this ..

There are 4 letters and 4 corresponding envelopes.in how ways can u put the letters into envelopes such that none of the letters are going into right envelopes

KinarSeptember 28, 2010 at 8:12 amwow @ this site, really better explained..

AnonymousOctober 3, 2010 at 6:27 amFinally…it MAKES SENSE. You, sir, are a wizard. *thumbs up for the humour too*

AnonymousOctober 3, 2010 at 5:13 pmWow man, I have a quiz on the logic of Permutations and Combonations sometime this week, and you have saved me. My book only goes as far to saying n!/(n-k)! and n/(n-k)!k!. You sir, are a lifesaver.

KalidOctober 8, 2010 at 10:03 am@Anonymous: Awesome, glad it helped.

peteOctober 7, 2010 at 5:30 pmhow maney combanations can i make out of the

the followig numbers and what are thay

5 7 12 13 14 36 26

Thank You notary pete @yahoo.com

OffendiOctober 16, 2010 at 9:23 amI like how, even when I’ve worked through the subject matter intuitively myself, your writing styles just makes it very pleasant to go over everything again.

One thing I still confuse over is eliminating possibilities (duplicates, for instance): when do you divide and when do you subtract? And what does it really mean when you’re doing either?

KalidOctober 16, 2010 at 9:35 pm@Offendi: Thanks — I find when I revisit topics I thought I knew, there’s even deeper insights lying there that I missed the first time around. It’s like reading a book more than once for a deeper meaning, I guess.

The division/subtraction question is something that still trips me up too. To me, I think of “figure out all the possibilities and remove the ones you don’t want”.

If what you don’t want is a few isolated, specific items, then you subtract them out. But sometimes what you don’t want is mingled with what you *do* what (i.e. every item has an evil twin or three that shouldn’t be there). In that case you have to divide them out. But, division is like subtraction: 15/3 = 15 – 5 – 5 (i.e. if you know each item has 2 evil dopplegangers).

VictorOctober 19, 2010 at 4:51 pmA box with 50 marbles (each marble is mark with 1 to 50 ). When a man pick a marble, lets say the number is 23, then the put back the number 23 into the box & stir the box, he trys again and the second time is still 23, 3th round still no 23 and 4th round still no 23.

Can you tell me what is the chances of of picking no 23 on the 4th round ?

Thanks.

AnthonyOctober 22, 2010 at 3:46 amMarco plans to give (not necessarily even) his eight marbles to his four friends. If each of his friends receives at least one marble, in how many ways can he apportion his marbles?

huirenOctober 26, 2010 at 3:33 pmtwo cards are drawn at random without replacement from a standard deck of 52 cards. what is the number of ways at least one ace can be drawn?

sindhuOctober 27, 2010 at 8:27 pmhi i have confusion when solving problems that is with restrictions say question is 0123…..9 how many 4 digits nos can be formed? 4,5 with out repeating these numbers

Edward KingOctober 29, 2010 at 1:23 amYou really helped me alot….

YunaLinaNovember 3, 2010 at 8:34 amHello and nice to met u. can u help me wit dis combination and permutation topic. the question goes like this. “The lock number on a bag consist of 3 digits. Each digit can be set from 0-9. The lock may be opened by selecting the correct combination of three digits. How many different lock numbers are possible if any repetition of digit is allowed and not allowed? thank you.

BushNovember 3, 2010 at 10:28 amThis is great!u r giving me hope of majoring in mathematics as a degree.thanks a lot.

KalidNovember 3, 2010 at 6:26 pm@Bush: Awesome, happy to help!

AJNovember 3, 2010 at 1:40 pmHi,

Help needed badly on this one!!!

How many ways can you split a box of 20 marbles into 5 groups (each having atleat 1 marble)?

Thanks,

AJ

MyraNovember 5, 2010 at 7:46 pmi am grateful for your website i just found it today. after close to 30 years i decided to go back to college and whew!!!! to say the least. but i finally got the idea. thanks. but i am still confused….with questions concerning a standard deck of cards…52 cards, the chances of getting 2 queens, 2 kings, 7 none face cards when you pick 5 randomly… do we count just the 4 kings that are possible??? this would really help me. by the way you are awesome!!!

KalidNovember 5, 2010 at 9:08 pm@Myra: Thanks for the kind words! Happy to help, though I’m not sure I understood the last part (2 queens, 2 kings, and 1 non-face card you mean?).

I have to break these problems down when doing them. Let’s start simple: what’s the chance of pulling 1 random card and getting a queen?

There are 4 queens, and 52 cards total, so a 4/52 chance of getting a queen.

And what’s the chance of getting another queen? Well, of the 51 cards left, there are 3 queens. The 2nd time we have 3/51 chance, and together we have a chance of

(4/52) * (3/51) = .004

How about pulling a king afterwards? Well, that’s 4/50. And another king? That’d be 3/49. So our total is

Q1 * Q2 * K1 * K2

(4/52) * (3/51) * (4/50) * (3/49)

And lastly, we have to pull a non-face card. We want 2 through 10 (ten cards), in any suit, so 40 cards total. Our chance of a non-face card is 40/48. So the final chance is

(4/52) * (3/51) * (4/50) * (3/49) * (40/48) = 0.00001846893

That is, unless I messed up my thinking :). This problem is definitely a bit tricky, I always have to start simple and work my way up. As you said, we actually need to consider all 4 kings since any of them can be used (the question is easier if we have to get a certain king, like the King of Hearts, since there’s less ways to make the hand). Hope this helps!

PriyaNovember 7, 2010 at 8:04 pmThis is how any right-brain thinker, say,

an artist, a farmer, or a “simpler human” would like maths to sound like – Kalid, u’re a “poetic” mathematician.

AnonymousNovember 8, 2010 at 12:28 amthe problem is that:how many 3 digit even numbers can i make with digits 1,2,3,4,5,6.its just that i’m not finding the way about it

DamiNovember 9, 2010 at 3:01 pmdo you have a tutoring bissness

KalidNovember 12, 2010 at 1:25 am@Dami: Not yet :).

Humphrey TugumeNovember 11, 2010 at 4:54 amwow i wish i was near u i would have given u a hug. THAT WAS GOOD AND SIMPLE TO UNDERSTAND.THANKS

KalidNovember 12, 2010 at 1:25 am@Humphrey: Haha, awesome! Glad it helped.

FranNovember 12, 2010 at 10:17 amHi, I’m writing an exam tommorrow, can you please help me with the following problem. I have to find the vale of n in the following equation:

21(nP4) = 7(nP5)

This is what I have thus far:

3n!/(4-n)!=n!(n-5)!

3n!(n-5)!=n!(4-n)!

3(n-5)!=(4-n)!

I would really apprieciate your help

Thanx!

JeremyNovember 16, 2010 at 12:40 am@Fran

This might be a little late, but it would be:

21(nP4)=7(nP5)

3n!/(n-4)!=n!/(n-5)!

3n!/(n-4)(n-5)!=n!/(n-5)!

*a little magic*

3/(n-4)=1

3=n-4

n=7

AnonymousNovember 23, 2010 at 5:29 pminhow many ways can the 12 members of volleyball team line up, if the captain and assistant captain must remain together

YaraNovember 25, 2010 at 1:29 pmYou’re really great Kalid :) Thanks for helping everyone out!

PradipNovember 27, 2010 at 9:02 amHi,

Given digits 2,2,3,3,4,4,4,4 how many distinct 4 digit numbers greater than 3000

can be formed?

I am confused in removind redundant 2s and 4s :(

AhadDecember 1, 2010 at 1:58 amThis is the kind of Blog that should be endorsed.It is a concrete method of analyzing problems with different respects.I thank each and every single person for experimenting the title with great acuity.(it is something that you and I will never learn in school) Thank you

KalidDecember 1, 2010 at 6:14 pm@Ahad: You’re welcome, thanks for the comment.

Sudipta sundar sahooDecember 2, 2010 at 5:49 amActually, the way it make understand, i think it is the best known method that anyone can understand .

GinaDecember 6, 2010 at 6:25 pmif 15 games are played, each with 2 teams, how many combinations are there to correctly choose the winner of each game?

deepakDecember 8, 2010 at 9:34 am@kalid

gr8 work.commendable.

@501

arrangement starts after placing the 1st person

2nd person can be placed either to the right or left of 1st person so 2ways

similarly for the 3rd person he can be arranged either to the left of 1st person or to the right of second person or inbetween them ,so 3 ways

…..pattern continues till 10 team members 10!

captain and asst cap should be treated as 1 group as they always stand together.arranging them is 2!

total combination =10!*2!

hope this is useful

TaylorDecember 8, 2010 at 4:24 pmThanks so much! This helped a lot. (:

KalidDecember 10, 2010 at 2:21 am@Taylor: No prob!

KrishnaDecember 10, 2010 at 10:57 amReally nice explanation. Now I don’t have to memorize the formula as I now understand how this works :-)

kafiDecember 13, 2010 at 10:31 amHey Kalid could you help me with a college level permutation question.

TX

kafiDecember 13, 2010 at 10:34 amIt would be very nice if could explain this question with explanation of permutation and combination.Hers’s the question:

Each of the 11 letters A, H, I, M, O, T, U, V, W, X and Z appears same when looked at in a mirror. They are called symmetric letters. Other letters in the alphabet are asymmetric letters. How many three letter computer passwords can be formed (no repetition allowed) with at least one symmetric letter?

And also what does this no repetition means.

Tx

jawadDecember 13, 2010 at 10:42 amDear sir,a great job from your side but i have simple suggestions for you to extend this article by adding the types of permutation,like 1)repetitions is allowed and 2)No repetition.

I hope you will consider this.

THnks.

FloDecember 29, 2010 at 6:24 amThanks, awesome material!

LucyDecember 30, 2010 at 9:21 amHey Kalid,

Thank you for you explanation above. However, as I was working with permutations/combinations, I really don’t understand the difference between these two problems.

There are 12 students in a class. Find the number n of ways that 12 students can take 3 different tests if 4 students are to take each test.

Find the number n of ways 12 students can be partitioned into 3 teams A, B, C so that each team contains 4 student.

The first answer was 34,650, and the second answer was 5925.

azraJanuary 1, 2011 at 2:04 ama student has 4 different pairs of shoes and never wears the same pair on 2 consequitive days.in how many ways can he wearshoes in 5 days?

caoimheJanuary 7, 2011 at 8:24 amhi ive a problem plse can you help! karen has seven subjects maths, french, german, history, english,geography, science.how many different ways can she do the exercises if 1. she must start with a language & 2. she must not start with a language.

Thanks

c.January 7, 2011 at 8:28 amhi ive a question plse help! 11 students 96 boys, 5 girls) are to line up in a straight line. In how many different ways can they be arranged if no student of the same gender may be next to one anither.

thanks

SumitJanuary 14, 2011 at 1:31 amWhat are restricted permutations and combinations? and What are complementary combinations? plz explain with example.

JoyJanuary 16, 2011 at 4:36 pmthis was the first blog i read in your site back in 2010, but didn’t comment then, do you know how i stumbled upon it – google.

i just googled permutation and combination, and betterexplained beat wikipedia in google in hits WOW !,

it was diffult to believe early on that a site can make maths so easy , i had a feeling that i wasnt smart enuf for maths, after encountering the boring maths books, but your site helped me gain confidence in my intellect and self in general. from then onwards i always look out for the intuition in the topics rather than the page with the formulas.

KalidJanuary 24, 2011 at 3:28 pm@Joy: Thanks — I’m really happy you were able to gain confidence, that’s really the core of it. A huge part of math is not the math itself, but learning how to “get” new ideas as they come along. It’s awesome you’re looking for intuition now!

MimiJanuary 26, 2011 at 8:39 pmI still don’t comprehend. But I know more than I did before :)

KalidJanuary 28, 2011 at 10:18 pm@Mimi: Happy even if it partially helped :).

KicabalesFebruary 4, 2011 at 6:40 pma building has 5 entrance gates and 4 entrance gates. in how many ways that a person can enter and leave the building?

can you please help me answer this question.. ASAP.. thank you..

jonalieFebruary 13, 2011 at 5:21 pmwew,, question: how many permutations can u make in the word ATLANTA… pls. help me answer. asap…tnx

AnonymousFebruary 19, 2011 at 5:48 amYour explaination and presentation gives a better and clearer picture thus making it easier for to explain to my students. Thanks again.

KalidFebruary 22, 2011 at 10:25 am@Anonymous: Glad it helped!

kamran alamFebruary 22, 2011 at 9:36 pmits better.not good

komolFebruary 25, 2011 at 11:02 pmthank u man, ur explanation did help alot…

Ninja MonkeyFebruary 27, 2011 at 10:09 amIs this permutation or combination? Please explain! Thanks!

Choose five books to check out from a group of ten.

AnonymousFebruary 28, 2011 at 10:28 pmOmg thxs so much this website is so helpful

AnonymousMarch 2, 2011 at 12:27 pmGreat job. Very helpful and well explained

KalidMarch 2, 2011 at 3:02 pm@Anonymous: Thanks!

cherryMarch 3, 2011 at 6:28 amReally nice explanation. Now I don’t have to memorize the formula as I now understand how this works :-)

KalidMarch 3, 2011 at 4:16 pm@Cherry: Thanks — it way better to understand than memorize :).

lucyMarch 7, 2011 at 3:46 amThanks a lot! I have two questions.

-1.There is a question to help me to explain my problem.(9%of men are color blind,if 4 men are chosen at random,find the probability that exactly 2 are blind)In this question we have to use combination:4C2.But I think that :

1st blind 2nd blind 3rd ok 4th ok

1st ok 2nd blind 3rd ok 4th blind

If we use combination (the order isn’t matter,so the two ways we have shown above are same(because the order isn’t matter)so may be wo may use permutation instead of combination.

-2How can we find if a question should use combination or permutataion? (if question does not tell us the order is necessary or not)

There are my 2 questions.Thanks~

aarrisMarch 8, 2011 at 5:22 ami have a question to all of you guys ….answer it if possible……one ice cream costs 10rs one biscuit costs 3rs and one toffee costs 50paise. i give you 100rs and ask you to buy a minimum of one item in each.i want hundred nos.all together and the cost also should be exactly 100rs. this problem has three solutions…tell me if u guys can…

RoxanneMarch 8, 2011 at 9:27 pmI did find my question in this blog, but no one has answered how to get to the answer.

A shipment of 12 tv sets contain 3 defective sets. in how many ways can a hotel purchase 5 of these sets and receive at least 2 of the defective sets?

i know nCr is used here as the order of the tv sets do not matter. The solutions say the answer is :

#2 defective + #3 defective

= 9C3.3C2 + 9C2.3C3

= 252 + 36

= 288

I dont understand this!!!!

ashok kumarMarch 11, 2011 at 10:57 amthanx !

it helped me and i think probablty is not that much difficlt as ppl think.

bababMarch 13, 2011 at 9:48 amit is really helping me for pssa test. thanx for this good and easy explanation

RohanMarch 18, 2011 at 8:12 pmThank you so much – this helped me alot!

KalidMarch 19, 2011 at 1:34 am@Rohan: You’re welcome!

lucyMarch 21, 2011 at 4:36 amThanks a lot! I have two questions.

-1.There is a question to help me to explain my problem.(9%of men are color blind,if 4 men are chosen at random,find the probability that exactly 2 are blind)In this question we have to use combination:4C2.But I think that :

1st blind 2nd blind 3rd ok 4th ok

1st ok 2nd blind 3rd ok 4th blind

If we use combination (the order isn’t matter,so the two ways we have shown above are same(because the order isn’t matter)so may be wo may use permutation instead of combination.

-2How can we find if a question should use combination or permutataion? (if question does not tell us the order is necessary or not)

There are my 2 questions.Thanks~

davidApril 1, 2011 at 5:56 pmif a flashlight, snack, water bottle and sweatshirt are packed. The sweatshirt needs to be packed first, how many ways can the items be packed.. I say 3*2*1= 6 but the answer key says 25… How do you get 25 ways??

Bryce OlsonApril 4, 2011 at 1:08 pmWow thanks a ton your a life saver

anynomousApril 12, 2011 at 1:22 pmyour a terrible explainer you didnt help at all

AnonymousApril 16, 2011 at 7:32 amAre these taken from Sams Series of Books?

KalidApril 16, 2011 at 6:44 pm@Anonymous: Nope, I just write ideas as they come :).

AnonymousApril 16, 2011 at 3:00 pmyou guys rock! Thank You so much know i can actually understand what i am doing awesome!!!!

KalidApril 16, 2011 at 6:43 pm@Anonymous: Thank you!

kashifApril 16, 2011 at 10:35 pmwhat a great thanks to you at this kindness in favour of students……… 4m pakitan

SamApril 17, 2011 at 6:21 amGreat article! But can you explain the difference in the problem I have now

What is Probability of 2 threes in a row followed by diff number on a 3rd roll?

What is probability of getting exactly 2 6’s on 3 rolls

For the first Probability of getting 3 is 1/6

and not getting 3 is 5/6

So we have 1/6 * 1/6 * 5/6 as answer

Am I correct?

Similarly for second we have same logic just in addition we multiply by 3!/2! Is that OK?

MattApril 17, 2011 at 1:22 pmThank you from the bottom of my heart. This site continues its tradition of nothing but the most well written articles. Thank you again and again.

AnonymousApril 18, 2011 at 11:44 amthank you

AnonymousApril 18, 2011 at 4:51 pmHow can you decide whether something is a per. or a com.?

For example: “How many different ways can a deck of 52 cards be arranged?” Is that a per. or a com.?

I utterly clueless when it comes to math, thanks in advance!

AnonymousApril 18, 2011 at 5:17 pmI have a question, it is

billy has 43 baseball cards, and scottie has 36. How many ways can Billy trade 2 of his cards for 2 of Scottie’s cards?

I thought this would be two stages with a conclusion that looked like

C(43,2) + C(36,2)

I just do not understand if I should be adding them together or multiplying, why I won’t, or if it is suppose to be permutation.

please help, so lost.

AnonymousApril 18, 2011 at 5:27 pmFor your deck of cards question it would be permutations because order does matter. Say you have a king of hearts and a queen of hearts…. then say you have a queen of hearts and a king of hearts… even though they are swamped it is the same hand of cards.

I hope this helps!

AnonnyApril 19, 2011 at 9:14 pmWOW, I love whomever wrote this!

This explanation was better than my teacher’s, to be quite honest…

This totally saved my life! (Or at least my next test grade…)

AnonymousApril 22, 2011 at 5:44 am1337

NikiApril 23, 2011 at 2:50 pmThanks so much! I was having so much trouble with these. By any chance, would you be able to show me an example involving people seated around a circular table? those types of questions always confuse me

KalidApril 24, 2011 at 12:16 am@Niki: Thanks, and great question. I’d like to make a post covering all types of combination / permutation problems and how to think about them.

AnonymousApril 23, 2011 at 10:11 pmHaha i love how you explain things. It gets interesting when it actually isn’t. THANK YOU :D

KalidApril 24, 2011 at 12:15 am@Anonymous: Awesome, I was hoping to make combinations bearable :). Thanks!

BashirApril 25, 2011 at 5:06 pmAwesome site Kalid!

Hopefully you can help me with problem thats been stumping me.

A box contains glass lenses for traffic signals. Five are red, Four are yellow, and three are green.

Find the probability of randomly selecting 3 lenses and getting red on the first selection, green on the second selection, and yellow on the third selection.

AnonymousApril 25, 2011 at 5:33 pmhey i just have a question that i dont get…

a lock combination comprises of 4 digits…

how many possible combinations are their in each case?

a)the digits can not repeat within the combination

b)the digits can repeat within the code

any help would be greatly apprecieted thx :)

AnonymousApril 25, 2011 at 5:38 pmThank you! this was very well written and thought out.

BashirApril 25, 2011 at 6:35 pmFor post #566

a) assume lenses are replaced

b)lenses are not replaced

KalidApril 26, 2011 at 1:34 am@Bashir: Let’s use a simpler example. Let’s say I have 2 red and 2 green lenses. What’s the chance of getting a single red lens? (2/4) = 50%.

Ok, now what’s the chance of getting a red then a green? Well, it’s 50% to get the red. If we pull it out, we have 3 lenses left. So there’s a 2/3 = 66% chance to get a green afterwards.

Therefore, the chance to get a red then green is 50% * 66% = 33%. Basically, you figure out the chances you need at each step and multiply. Hope this helps!

sissyApril 26, 2011 at 11:02 amI need the list of all possible combinations for a 5 digit number using 2 and 5. Repeats are allowed

KiraApril 26, 2011 at 7:15 pm^ It’s easy:

_ _ _ _ _

Since order matters:

2P1 = 2 pick 1 ( 2 # pick 1 of them)

2P1 * 2P1 * 2P1 * 2P1 * 2P1

OR you can say it as:

2^5

Math haterApril 26, 2011 at 9:04 pmHi khalid.can u please explain me the solution for this problem

The simplistic language has only 2 unique values and 3 unique consonants.every noun in simplisatic has the structure cvcvc where c stands for a consonant and v stands for vowel .how many diff nouns are possible in simplistic._

I know this is a combinatorics problem but don’t know how to solve it..pls help

Math haterApril 26, 2011 at 9:12 pmIf u can pls explain this I would be cleared about other similar problems in future .thanks in advance.the page is helpful specially when u never studied the probabilities and bow you have to for the GMAT..

KiraApril 28, 2011 at 3:47 pmAssuming an alphabet of 26, 5 vowels and 21 consonants, and also assuming you can repeat letters that you already picked, then:

21P1 * 5P1 * 21P1 * 5P1 * 21P1

or if you wish, 21 * 5 * 21 * 5 * 21

or, (21^3) * (5^2)

or, 231525 nouns.

The reason why it’s like that is because I assume that you can pick let’s say the letter T for the first consonant, and then T again for the second consonant; the choices of vowels and consonants don’t decrease for every letter you picked. You can use the letters again and again.

MiraApril 29, 2011 at 6:32 pmThx for helping me understand such a complicated puzzle and my project…..:D

Math haterApril 30, 2011 at 2:17 pmThanks for the explanation Kira…I get stuck with these problems alot I need to check in the answer again.thanks anyways.I dread these kinda problems.rest I can do well.but permutations and combinations takes away all my confidence.

Math haterMay 1, 2011 at 8:36 amHere is another problem.in how many ways can te letter abacus can be arranged such that always appear together .

I get that there are 6 letters 3 of them which are vowel 3 vowels can be arranged as 3!/2! as a appears twice but I m getting 6! As the consonant can be arranged but the answer to this is 4!3!/2!.what I m doing wrong cab somebody explain.thanks in advance

AmnaMay 1, 2011 at 10:38 amAre there any specific words to identify what to apply in a question? Permutation or Combination?

Math haterMay 1, 2011 at 6:43 pm@Anna when order of an event matters then it’s a permutation when order doesn’t matter it is a combination.for ex:In how many ways I can give 3 tin cans to 8 people.here order doesnt matter.the explanation for this is provided at the top..however recognizing the problem does help alot but it really matter to solve it correctly..combinations are easier than permutations..

faryalMay 4, 2011 at 4:20 ami found this site good

PrasannaMay 6, 2011 at 12:32 amCan’t get any better the way you explained, thanks.

osamaMay 6, 2011 at 2:46 pmA student is to answer 10 out of 13 question:

1.How many if he must answer at least 3 question from first five question ?

dixit sharmaMay 8, 2011 at 5:43 pmthanks a lot… this has been a wonderful help to me…

Joshua A. DIggsMay 9, 2011 at 5:28 amI would like to know how did the formular came about and how did this formula came about

help meMay 9, 2011 at 6:46 amhi.. can anybody help me with this question..

john has 7 best friends. his brother is getting married and he plans to invite them to the wedding. in how many ways can this be done if:

a)he wants to invite one or more of them to the wedding

b)mary and suzy must be invited

c)mary is invited then suzy cannot be invited

jewlieMay 10, 2011 at 7:32 amhi my name is jewlie and this is really difficult to read yu should make it esier and guess what!! i like to chew gum

AnonymousMay 17, 2011 at 4:43 pmplease make formulaes clear i dnt understand combinations

george chikosMay 18, 2011 at 1:31 ami find this exciting can you pliz assist me with question

how many ‘words’ each consisting of two vowels and three consonants canbe formed from the word “integral?”

KatieMay 20, 2011 at 10:10 amI am really confued as to how to figure this out and what formula to use…… “you are scrambling the letters to the word Mississipi, how many different arrangements can you make?” if you could help me out that would be amazing because my math grade depends on this stuff lol

Bon CrowderMay 21, 2011 at 4:22 amI did a video on it here: http://mathfour.com/finite-math/permutations-and-combinations-how-to-tell-the-difference

KateMay 22, 2011 at 11:16 amIn answer to comment 561:

In round table questions it is always one less than the number of people sitting around the table that is the factorial number. This is because we have to fix one person in order to make the answer finite. Therefore if there are nine chairs at a table the number of ways that they can be filled is 8!. This is true unless we are talking about perhaps fixing a husband and wife sitting together at the table and then we will use them as the fixed pair – so we’d say that there were 8! x 2 ways that they could sit together (as they could sit each side of each other). Hope that this helps – it took me ages to get used to the idea!

Shehryar KhanMay 24, 2011 at 11:11 amThank u a lot finally this makes sense u are better than my teacher at explaining this

censoredMay 24, 2011 at 6:03 pmHow many two person committees can be chosen from a group of eight people?

AnonymousMay 24, 2011 at 8:22 pmthank you so much my final is tomorrow and im just now getting it :)

KinyJune 4, 2011 at 11:21 pmPlz answer these

if 12 persons are seated at round table what is the probability that two particular persons sit together?

stacyJune 5, 2011 at 6:34 pmHow many ways can you choose four toppings for a pizza if there are eight toppings to choose from?

AnonymousJune 6, 2011 at 8:14 amHow Many combination can I get out of 31 numbers using 6 number at a time?

Martin WilsonJune 6, 2011 at 8:35 amHow many combination can 1 to 31 change up using 6 numbers.

kateJune 6, 2011 at 4:02 pmdude i still don’t get permutations

heheheheJune 6, 2011 at 4:03 pmi’m # 600!!!

AnonymousJune 12, 2011 at 7:27 pmA flush contains all five cards of the same suit. Find the probability of being dealt a flush in hearts from a standard 52 card deck

CraigJune 12, 2011 at 7:31 pmSomeone answer #601 I have the same problem

Philip StoneJune 12, 2011 at 7:32 pmI have the same question as 601.

AnonymousJune 12, 2011 at 7:32 pmme too. help with 601

THANKFULJune 17, 2011 at 7:28 amTHANK YOU SO MUCH!!

You explained it better than my teacher!!

It was zoo confusing at first but now i totally understand! :D

THANKS AGAIN :)

KalidJune 21, 2011 at 8:52 am@Thankful: Awesome, glad it helped! :)

Denis B. KipkorirJune 21, 2011 at 1:15 amThanks!You got it right. This is more helpful. I understand know

j glassJune 23, 2011 at 12:52 amyou did a better job explaining than Sal at Kahn Academy. That is a big compliment!

KalidJune 23, 2011 at 8:33 am@j glass: Thanks, glad it was helpful!

Why I love Internet « InundatedJuly 8, 2011 at 10:49 am[…] how to blog, know thyself. Math-o-phobic may find some solace here after reading articles like Permutations and Combinations. The aha-moments that the site tries to get from the article readers of a post is something that […]

aravindJuly 9, 2011 at 11:15 pmfantastic :)

FRANCIS ALEGOJuly 11, 2011 at 4:23 amhow many arrangements can be made from 3 letters chosen from the word PEAT if the first letter is a vowel and each arrangement contains 3 different letters?

FRANCIS ALEGOJuly 11, 2011 at 4:27 amthe explanations r just wow.i liketh.THANX BIG

johnJuly 11, 2011 at 1:07 pmhelp me with this question pls:Raila has 7 different posters to be hanged in her bedroom,living room and kitchen.Assuming she has plans to place at least a poster in each of the 3 rooms,how many choices does she have?

IvanJuly 13, 2011 at 3:04 amHello, are you still managing this site?

I have a question.

A quiz consists of 10 true/false questions. How many different answer sheets can be obtained?

Answer : 1024

But, I don’t know what is the working. 2^10? Why?

IvanJuly 13, 2011 at 3:22 amIs it like #571?

Dennis DJuly 15, 2011 at 12:35 pmA financial advisor offers 8 mutual funds in the high risk category, 7 in the moderate risk category, and 10 in the low risk category. An investor decides to invest in 3 high risk funds, 4 moderate risk funds, and 3 low risk funds. How many ways can the investor do this?

BrittneyJuly 22, 2011 at 3:40 pmHelp, this is for practice for my stats class and its been a while since I had done these types of problems. There are 11 different statistics books, 6 different geometry books, and 3 different trigonometry books. A student must select one book of each type. How many different ways can this be done?

AnonymousAugust 2, 2011 at 6:09 pmwow…………

i can get 4 flat for ma basic discrete maths

AnonymousAugust 5, 2011 at 5:23 pmLoved your explanations. I was having trouble with my summer assignments and I found this extremely helpful.

timAugust 6, 2011 at 12:46 pmanyone have a cash 5 qp lottery ticket in numerical order 1 thru 5?

pratishthaAugust 7, 2011 at 10:12 amcan uh plz..give me a real life example in which both permutation and combination are used?…plz its urgent..:(

saiAugust 9, 2011 at 6:35 amit is too difficult

NANDEESHAugust 13, 2011 at 9:02 pmHi

Suppose there are some envelopes with addresses written. There are same number of letters which are to be placed in the envelopes.

If we place the letters in the envelopes randomly there is only 1 chance out of n!ways of putting the letters into envelopes.

If no letter matches with the envelope, such a permutation is called ‘derangement’.

If there are 2 envelopes A & B and the letters are ‘a’ & ‘b’ the derangement is(b,a). That means letter ‘b’ is placed in envelope A and letter ‘a’ is placed in envelope B.

If there are 3 envelopes A, B & C and the letters are ‘a’,’b’ & ‘c’ the derangements are (b,a,c) and (b,c,a).

For the case of n=4, there are 9 derangements.

For the case of n=5, there are 44 derangements.

=========

Take another case:

A certain number of husbands are standing behind a screen. Their wives come and stand in some random order in front of the men. The screen is lifted. Lo! Wives and husbands have got mixed up. There may be some consolation. The chance of at least one wife standing in front of her husband is about 64%.

The number of derangements are about 36% of all the permutations.

Thanks for reading.

Kalid can explain derangements in a better way.

KalidAugust 14, 2011 at 9:17 pm@NANDEESH: Awesome! I hadn’t heard of the term derangement before, I like the concept :). Thanks for explaining it.

AnonymousAugust 15, 2011 at 1:12 amHi,

I myself had not heard of derangements two days ago. I was given a problem to solve. It involved 4 different color balls to be put in 4 containers of same four colors such that no ball will be placed in a box of same color. I wrote all permutations and got the answer. Then I calculated the number for the case of 5. I lost my sleep wondering how many it would be for n=6. I tried many formulae but failed. I called it x(n).

However I discovered that x(n)=(n-1)*(x(n-1)+x(n-2)).

I searched in wikipedia and to my ecstacy I found a similar formula.

I learnt that no.of derangements is denoted by !n and called subfactorial n.

Somewhere, I learnt that !n and n! are related as below:

!n=floor(n!/e +1/2)

(Floor means ignore the decimals.)

I patted myself for some original thinking on a topic which is about 300 years old!!! OK, I am 300years late. So what!

Honestly I had not heard of derangements in school or engg college. But I am happy now.

NANDEESHAugust 15, 2011 at 1:13 amMy comment on derangements is appearing under Anonymous.

Sorry.

NANDEESHAugust 15, 2011 at 1:40 amHi,

I myself had not heard of derangements two days ago. I was given a problem to solve. It involved 4 different color balls to be put in 4 containers of same four colors such that no ball will be placed in a box of same color. I wrote all permutations and got the answer. Then I calculated the number for the case of 5. I lost my sleep wondering how many it would be for n=6. I tried many formulae but failed. I called it x(n).

However I discovered that x(n)=(n-1)*(x(n-1)+x(n-2)).

I searched in wikipedia and to my ecstacy I found a similar formula.

I learnt that no.of derangements is denoted by !n and called subfactorial n.

Somewhere, I learnt that !n and n! are related as below:

!n=floor(n!/e +1/2)

(Floor means ignore the decimals.)

I patted myself for some original thinking on a topic which is about 300 years old!!! OK, I am 300years late. So what!

Honestly I had not heard of derangements in school or engg college. But I am happy now.

DinaAugust 17, 2011 at 9:38 pmHi NANDEESH.. Nice work.. It seems interesting..

AnonymousAugust 18, 2011 at 5:24 pmgood work

ajayAugust 24, 2011 at 5:44 amhow many diameters can be drawn from 21points on a circle?

AnonymousAugust 24, 2011 at 11:27 pmnice explanation

AnonymousAugust 25, 2011 at 3:46 pmOMG THANK YOU SOOO MUCH!!! My teacher is soo confusing its crazy if i hadnt found this site i would hav failed my quiz tomorrow lol :) again thank you!!!

KalidAugust 25, 2011 at 3:59 pm@Anonymous: Thanks! Glad it helped.

LinkAugust 28, 2011 at 12:36 pmI can’t believe how long this thread has been running!

My question. My Dad and his buddies (12 in all) go on a golf outing each year. They play 3 or 4 rounds (so I need to calculate for both situations) during the trip and he often ends up playing with the same players and never plays with others. He wants me to come up with a set of foursomes for these rounds that helps them play with the most golfers over the duration of their trip.

Can you help me choose his foursomes?

NANDEESHAugust 29, 2011 at 8:43 pm@Link

If I have understood your problem correctly, one possible algorithm is as follows:

1. Assign random numbers to the 12 players. (Like Rand() in excel)

2. Sort them in the sequence of random numbers.

3. Split the list into 3 or 4 groups as you like.

4. Repeat the process whenever you want to change the grouping.

linkAugust 30, 2011 at 7:15 pm@Nandeesh – I think you’d be surprised at how many duplicates and partnering potential you would miss by completely randomizing it. It takes a very precise placement to optimize both. Give it a shot and you’ll see what I mean.

Thanks for trying.

NANDEESHAugust 30, 2011 at 9:40 pm@link

Can you pl. illustrate the issue with examples?

This site is for explaining issues and solutions better. If the problem is not stated properly how can you expect a solution to your satisfaction?

sweetaAugust 31, 2011 at 6:59 amThanks a lot, superbly explained… :)

linkAugust 31, 2011 at 2:40 pm12 Golfers play 3 rounds in groups of 4. They need to know how to group the teams during each round so that each player maximizes the number of other players he plays with.

For example:

Round 1: T1: 1, 2, 3, 4 T2: 5, 6, 7, 8 T3: 9, 10, 11, 12

Round 2: T1: 1, 5, 9, 2 etc.

Obviously, it is necessary in the above example for player 2 to play with some player with which he has already played. The goal is to minimize these duplicates and prevent players from missing out on playing with as many different players as possible.

Two options for solutions: 1. It would be ok to have this problem of combinations explained and a pattern developed so I can solve this one problem. 2. I would also entertain programming code to help to solve this problem using different criteria in case they have a different number of players or rounds of golf in future years.

Thanks and hope this helps.

KalidAugust 31, 2011 at 3:11 pm@link: Really interesting question. I don’t know if this is a perfect solution, but it gives really good results.

Write your numbers out in a circle. Luckily, for 1-12 you can just make a “clock” :).

Start with #1, and make a rule to get 3 other people in the group. This rule might be

(me, me + 3, me + 6, me + 9)

which yields (1, 4, 7, 10). This is one group. Now, run the rule for the next available person (#2), so you get another group of (2, 5, 8, 11). Do the same for the next available (#3) and you get (3, 6, 9, 12).

Ok. That’s round 1. For round 2, start with person #1, and create a *new rule* to make a group he hasn’t seen before, such as (me, me + 1, me + 4, me + 7). This gives (1, 2,5, 8 ) and (3, 4, 7, 10) and (6, 9, 11, 12).

And… you guessed it. For the next round, create *another* rule that gives a new group, such as (me, me + 2, me + 5, me + 8 ). This gives (1, 3, 6, 9) and (2, 4, 7, 10) and (5, 8, 11, 12). [If you can’t finish the rule, just use the available people].

In summary:

Round 1: (1, 4, 7, 10) & (2, 5, 8, 11) & (3, 6, 9, 12)

Round 2: (1, 2, 5, 8 ) & (3, 4, 7, 10) & (6, 9, 11, 12)

Round 3: (1, 3, 6, 9) & (2, 4, 7, 10) & (5, 8, 11, 12)

My intuition is this optimizes for person #1, but in the process mixes up everyone else as best it can. You can see that “4, 7, 10” play almost the same in round 2 and 3, so a manual mix at that point might be good.

The neat thing is this can be done on paper (I just did it by drawing out the clock and coming up with a few rules :)).

linkAugust 31, 2011 at 6:05 pmHey Kalid – Thanks for the time you put into that. For drawing a circle you did amazingly well. However, the circle/rule method you chose created 31 “misses” – meaning that 31 times someone missed playing with another player. For comparison, someone had written a paper on this “social golfer problem” (I didn’t know it had a name until I started looking for an answer) and he produced only 18 misses. However, there were 3 players on his suggestion that played with another player in all three rounds. Your solution put 5 players in that position.

I wrote a computer program to try to solve the problem and created no “3s” (as a rule, I wouldn’t allow it) but also created 23 misses. So, I’m not sure exactly which is best. my program limited the number of times you repeated playing with another golfer to only 2 but had 5 more misses.

Either way, I think I’m done with this problem. But, I thought your crafty solution deserved a “attaboy!” Thanks!

NANDEESHAugust 31, 2011 at 11:31 pm@link

SOCIAL GOLFER PROBLEM

==================

One solution for the given problem is given below.

First the terminology:

===============

The players are numbered 1 to 12.

They are to be placed into three groups A,B and C.

For group A, the next group is B and the previous group is C.

For group B, the next group is C and the previous group is A.

For group C, the next group is A and the previous group is B.

Now the iteration process:

===================

Let the groups for the first round be A=(1,2,3,4),B=(5,6,7,8) and C=(9,10,11,12).

For the next round:

1. First player of each group remains in the same group.

2. Second player moves to next group.

3. Third and fourth players move to previous group.

Repeat this method 4 more times.

So, second round looks like A=(1,7,8,10),B=(2,5,11,12) and C=(3,4,6,9).

Third round looks like A=(1,11,12,4),B=(7,2,6,9) and C=(8,10,5,3).

Fourth round looks like A=(1,6,9,10),B=(11,7,5,3) and C=(12,4,2,8).

Fifth round looks like A=(1,5,3,4),B=(6,11,2,8) and C=(9,10,7,12).

Now, reset the starting grouping.

So, for the 6th round, A=(12,6,7,8),B=(1,2,4,5) and C=(3,9,10,11).

Iterations like earlier, for the next round:

1. First player of each group remains in the same group.

2. Second player moves to next group.

3. Third and fourth players move to previous group.

Repeat this method 3 more times.

So, seventh round looks like A=(12,1,3,11),B=(2,9,6,8) and C=(10,7,4,5).

Eighth round looks like A=(12,2,10,5),B=(9,7,1,11) and C=(4,3,6,8).

Ninth round looks like A=(12,9,4,8),B=(7,3,2,5) and C=(6,10,1,11).

Thus, in a total of 9 rounds each player would have played with every other player.

By this method, misses have been avoided. But whether duplicates are minimized, I am not sure.

NANDEESH

KalidSeptember 2, 2011 at 9:46 am@link: Ah, no fair, you’re giving me research topics! :). It is a really interesting problem though, and at first glance I thought some combination/permutation trickery could solve it. But it looks like there’s a more complicated algorithm that’s needed. Really cool question though!

@Nandeesh: Thanks for the input — I think in this case, however, we can only have 3 or 4 rounds.

NANDEESHSeptember 2, 2011 at 9:42 pm@link

Pl. let us know the minimum number of rounds in which each of 12 players would have played with every other player in groups of 4.

The number cannot be as less as 3 or 4. Because, a player can cover 3 different players per round and so to cover 11 other players, he needs 4 rounds. There need to be additional rounds for others to do the same.

9 rounds is a solution. But I am sure, it is on the higher sde.

Quite an interesting problem. I liked it.

SherzSeptember 6, 2011 at 3:43 pmO! My…. thankxxxx i actually understand thank yew!! Xo

KalidSeptember 6, 2011 at 7:44 pm@Sherz: Awesome, you’re welcome!

DidierSeptember 8, 2011 at 7:46 amWe have 6 players (numbered #1 to #6) and would like to make 2 teams of 3 players. What are the combinations? Thanks in advance for your help.

AnonymousSeptember 11, 2011 at 1:05 pmthank you so much! finally i’ll know what i need to know for the big test and will be able to finish my homework! thanks! :)

KalidSeptember 11, 2011 at 11:59 pm@Anon: Awesome, glad it helped!

JACKIESeptember 16, 2011 at 11:52 amholy macaroni, everything makes so much more sense :)

KalidSeptember 16, 2011 at 12:34 pm@JACKIE: Thanks, happy it helped!

PrasannaSeptember 17, 2011 at 4:23 pmWonderfully explained….Relay helped me…Thanks

KalidSeptember 18, 2011 at 7:46 pm@Prasanna: Thanks!

BotaSeptember 26, 2011 at 9:55 pmThank you very much, it’s really the best explanation!

aditya shahSeptember 27, 2011 at 2:23 amcan you please solve this for n : (n+1)! = 42 (n-1)!

shivohamSeptember 27, 2011 at 3:15 pmvery good

eminemSeptember 27, 2011 at 3:17 pmi like the way you demonstrated !

johnOctober 1, 2011 at 1:45 pmIf there are 6 runners in a 100-yard dash. How many ways are there for three medals(gold, silver, bronze) to be awarded if ties are possible?

DeoOctober 6, 2011 at 1:31 amHow to obtain the the simple ramdom samples of size two from the listed members below?

members: A,B,C,D,E and F

DIvine BateOctober 8, 2011 at 8:19 amPlease help me on this problem: In how many ways can six out of ten beads be arranged round a ring

danOctober 16, 2011 at 10:36 amcan someone confirm 64 is correct for the following please?

6 football teams are playing each with a possible out come of win lose or draw, is there really only 64 combinations to make sure i get the right answer?

thanks Dan

rock4realOctober 27, 2011 at 1:00 amPLS, can any one help me solve this question?

there are 10 books on a shelve, but the blue covers of 2 of them are to gothere and they must not be put together. how many ways can the books be arranged so that the red covers are apart.

stupidheadOctober 28, 2011 at 9:37 amwonderful job explanening to a 5th grader

MNovember 2, 2011 at 5:15 pmHi! I was wondering if you could show me how to figure out howmany permutations there are in the word Halloween

gideon jumaNovember 16, 2011 at 7:24 amhelp me with this question;

a password consist of two letters of the alphabet followed by three digits chosen from 0 to 9 repeats are allowed how different possible passwords are there?

the mad 1November 17, 2011 at 8:57 amwow !!!!!!u rock who evr u r thanx can u pls help me trig 2 pls pls with sugar in top pls

WEBSTAR LUNGAFANovember 17, 2011 at 11:36 pmTanx a lot 4 aidin us,@ first it was hard but now i av a clue

ALSHABAABNovember 18, 2011 at 2:00 amI LIKE MATHEMATICS

edoggNovember 21, 2011 at 6:40 pmHello….I see you solve everyones problems, maybe you can help me out. I need to pick 8 winners of 8 football games….but not just who wins the game, they have to cover the spread. How many combinations are there? thanks!

NJERU DANNovember 23, 2011 at 3:16 amwow u hv got us from dark then i love maths

jhynx_23November 24, 2011 at 10:29 pmwhoah..diz site really help me a lot and all the students out there!!!

Now I really love mathematics.

Thankzhieee!!!

jhynx—a BSEd Math major. xD

KayeNovember 26, 2011 at 11:17 pmThanks for this. This is a big help since my reference book is not detailed as this which makes me browse in the library and so. </3

KayeNovember 26, 2011 at 11:21 pma password consist of two letters of the alphabet followed by three digits chosen from 0 to 9 repeats are allowed how different possible passwords are there?

26*26*10*10*10 = 676,000

:)

As I website possessor I believe the content material here is rattling great , appreciate it for your efforts. You should keep it up forever! Best of luck.November 28, 2011 at 2:42 amA few months ago the mabezat virus created havoc on on three of my clients servers.I immediatly contacted Trend, who told me to download the latest patern(How friking stupid do they think I am)I eventualy had a guy from Trend sitting in my office trying to resolve the problem, he could not even sort it out but promised that a bandage patch would be released the following day.Needless to say, my high priority client had already been down for 48 Hours, I gave Trend a ultimatum to have a solution working soulution in place within 2 hours or lose the business.

HaveaQuestionNovember 29, 2011 at 3:34 pmcan you help me with this question? —> it is really confusing to me, and it will be a really big help for me if you explained :) Emily’s school offers 3 English classes and 4 History classes for her to choose from. She must choose 3 of these classes to complete her schedule. If exactly one of these must be an English class, how many different combinations of classes are possible for Emily?

nskNovember 30, 2011 at 6:50 am3English , 4 History class

you need to select 3 classes

for each selection at least one should be english

so there are 3 cases

1) the classes you selected are all english = 3C3

2) 2 english + 1 history = 3C2 + 4C1

3)1 english + 2 history = 3C1 + 4C2

add 1)+2)+3) will give answer

Please check if its correct

John L. FerriDecember 5, 2011 at 7:29 amMost MP3 music player can play in several modes: serial, shuffle, and random. Serial mode plays the songs in order. Shuffle mode creates a randomly ordered list and plays through the list so each song is played one time. Random mode plays a random song each time so that repeats are probable (selection with replacement.) If 10 songs are available, what is the probability of all 10 songs being played if 20 plays are listened to?

how to study effectivelyDecember 8, 2011 at 8:16 pmYou can definitely see your enthusiasm in the work you write. The sector hopes for more passionate writers such as you who aren’t afraid to mention how they believe. Always follow your heart.

CarrieDecember 10, 2011 at 6:24 pmmakes sense, yet I got a question wondering if you can help me solve because it seems it is uding both?

Four different mathematics books and six different physiology books are to be arranged on a shelf. How many different arrangements are possible if

the books in each subject must stand together (permutation since order matters) But do i multiply the answers together? IE 4*3*2 * 6*5*4*3*2 or add 4*3*2+6*5*4*3*2

second part of questions is if only the mathematics books stand together? then I have no idea! Help!?

Teg599992December 12, 2011 at 6:19 pmI WAS FEELING SO STUPID WHEN I BEFORE I GOT ON THIS WEBSITE NOW I DONT ANYMORE IM HAPPY I NOW KNOW HOW TO FINALLY DO PERMUTATIONS AND COMBINATIONS AND NOT LOOK LIKE A FOOL IN FRONT OF MY CLASS(AP). YOU SAVED MY GRADE IN MY ALGEBRA CLASS. SO, I WANT TO THANK YOU SO MUCH.

EvaDecember 14, 2011 at 10:20 amYOU’RE SOUUND :D

Simon BridgeDecember 17, 2011 at 12:13 amThe next logical step from above is – you give those 8 people 3 contests and award the winner of each with a can. In this case, it is possible for one person to win more than one prize. But we don’t care the order of the prizes. (If this has already been pointed out I missed it.)

I could say that 8x8x8 is the total number of ways to distribute the prizes, and divide out the dupes (AAB = ABA etc) or I can realise this is the 3 single-prize count (which we got) + the 2-prize count (where one of the prizes is 2 cans) + the 1-prize count (the prize is three cans – luck-ee) which is 8.

But I want to derive the “combination with repeats” formula.

@Carrie: in the first version, you use combinations on each of the book types – since they can be disordered amongst themselves – add them up: that’s how many ways to stack them in one order. Multiply that by the number of different ways to order the subjects on the shelf.

Carlie JukichDecember 20, 2011 at 9:28 amPretty impressive posts, thumbs up for the great work.

kalidDecember 20, 2011 at 9:21 pm@Carlie: Thanks!

aishaDecember 21, 2011 at 3:01 pmList all the possible ways that the word GREAT could be arranged. How many possibilities did you find? Work out this same problem using the formula for permutations that you learned in this lesson. Did you miss any in your original count? Explain why the formula of permutations is helpful for most problems.

christine :)December 23, 2011 at 12:25 ami already printed the comments up to post # 667… it helps me a lot, as a BSED-MATH STUDENT… ty so much… but, im still confused on the four letter combination from the word OUTLOOK… pls help me… especially kalid.. :))

christine :)December 23, 2011 at 12:31 ami already printed the comments up to post # 667… it helps me a lot, as a BSED-MATH STUDENT… ty so much… but, im still confused on the four letter combination from the word OUTLOOK… pls help me… especially kalid

R.magdellaDecember 24, 2011 at 7:54 amsir, i did a lot of question by the lesson PERMUTION AND COMBINATION but one thing i don’t have any idea to solve about this question,

( LCM)- 4!,5!,6!

I request you to understand me about it.

Saddam DeenariDecember 27, 2011 at 10:28 pmthat was nice explanation, especially defrentiating the combination by group and permutation by list………… thanxxxxxxxxxxxxxxxxxxxxxx alot

christineDecember 28, 2011 at 2:11 amkalid??

JeffreeyDecember 29, 2011 at 12:41 pmSir pls explain for me this problem

A box of one dozen eggs contain one that is bad. If 3 eggs are chosen at random what is the probability that one of them will be bad ?

How the answer comes 0.25 pls explain for me

SultanJanuary 4, 2012 at 5:11 amIt was really helpful and a nice way to differentiate between Permutation and Combination :)

AngeloJanuary 4, 2012 at 11:16 pmHello Jeffreey, I am not sure if you did managed to understand your question, but if not, I hope this will help you!

First off, lets just focus on the 12 eggs and forget about the bad egg. The question we want to solve first is how many different sets of three eggs are there? Inorder to solve this we use C(12,3). Thus, C(12, 3) = 220. Therefore there are a total of 220 sets of eggs if we were to choose 3 eggs at random.

Now we need to calculate within those 220 sets, how many of the sets contain the bad egg. Inorder to solve this we set the following up. Label each of the eggs. First egg is E1, second egg is E2, and so on, and you would have E1, E2, E3, E4, …, E12. Now we choose the first egg to be the bad one. Hence we would have the following sets: (E1, E2, E3), (E1, E2, E4), (E1, E2, E5) and so on. Notice that there are 11 options for the second egg, and 10 options for the third egg for each set. Since order does not count we can now calculate C(11,2). Thus, C(11,2) = 55. Therefore, there are 55 sets of three eggs that have the bad egg.

Prob=55/220 = 0.25 or 25%

I hope this helped you!

:)

zarirJanuary 11, 2012 at 7:21 pmPlease explain me why you multiply the choices when doing permutation, not addition, or subtraction, or division?

zarirJanuary 11, 2012 at 7:22 pmplease let me know why you multiply the number of choices when doing the permutation. Why not addition?

Ignore my previous comment. The email address was not right there.

r3xJanuary 12, 2012 at 3:44 amIn response to [email protected] more of a physics problem you know, and is best solved by relativity.. but since you asked:-

[A.]D__________________________<[.B]

*dist. b/w trucks is 75(not mentioning units as length here is always in miles and time in min and secs as mentioned)

As both trucks are approaching each other the time to their collision is directly proportional to their speeds, hence making one of the trucks(B) a stationary point 75miles away from A and A is trying to race to B.

Now, adding their speeds(as we just screwed up the velocity with relativity) we get (11/12 + 16/12)miles/min or if said simply, for A, 11 miles per 12 min & for B, 16 miles per 12 min, so, if you try to put some head into it, your problem says 1min before they collide, which is similar to 1min before A reaches B.

Hence You use [Distance=time*speed] you'll get =}

D=(11/12 + 16/12)*1(as the speed was computed for 1min)

You get dist=27/12=9/4=2.25miles

^Stupid question really..

r3xJanuary 12, 2012 at 3:54 amIn response to [email protected] possible to add for very small permutation examples, eg-in the case of tossing two coins one silver and one copper, you’ll get 4 permutations as order does matter here and you can simply imagine something so easy even if your a downy, but for suppose the balls on a pool table you’ll have quite a hard time thinking out all the various mutations unless of course your a born Einstein like prodigy or something :P

Whereas in combination its more of a common noun case unlike the proper name problems in ordered mutations which makes it that much more easy going. ;D

r3xJanuary 12, 2012 at 4:23 amIn response to [email protected](cool name :P)-Its a one liner really, in such cases you calculate the total probability of the egg drawn not being bad in all three trials and subtract it from Max probability ie-1(as the 12 eggs combined make up an exhaustive event).In this case, similarly, =1 – 11/12*10/11*9/10 = 1 – 9/12 = 1 – 0.75 = 0.25//xD

.

Ps-If your so under-informed that you dont know what are the three fractions;

=} 11/12 (starting probability of drawing an intact egg)

=} 10/12 (probability after you’ve already failed at drawing a bad egg and chose a normal egg)

=} 9/12 (probability ——“——– and chose a normal egg TWICE!!)

r3xJanuary 12, 2012 at 9:53 pmIn response to [email protected]-Its pretty much a basic math problem, so, as you know LCM put simply is the multiplication of all the prime numbers(in pairs or without) that make up the numbers in the problem whose LCM is to be taken, and you might know that x! means 1*2*3*4*5*………….*x , using the same property here we have;

4! = 1*2*3*4

5! = 1*2*3*4*5

6! = 1*2*3*4*5*6

Hence here, as you might notice after sorting the numbers to be used in the LCM, its 6! ie 1*2*3*4*5*6 = 720//xD

r3xJanuary 14, 2012 at 4:32 amIn response [email protected] As simple as it goes :)

Case I – If your not replacing the book that’s already been chosen, then it’ll be =}

=11/20*6/19*3/18

=11/380 = 0.0289474

Case II – If your picking a book then putting it back in, then it’ll be =}

=11/20*6/20*3/20

=198/4000

=99/2000 = 0.0495

r3xJanuary 14, 2012 at 8:41 am^700 comments.. HECK YEAH!!!

:putsonsunglasses:

YahzidJanuary 25, 2012 at 10:30 amI teach african drums and I have 4 drum beats and a rest. how many different patterns can I create. there are only 4 spaces to choose from. The answer is 4 patterns but I got the answer from playing all the possible drum patterns. how do get the answer using the formula.

kalidJanuary 25, 2012 at 11:23 am@Yahzid: Good question. Basically, you are choosing 1 beat of the 4 to have a rest come afterwards:

1 Rest 2 3 4

1 2 Rest 3 4

1 2 3 Rest 4

1 2 3 4 Rest

The formula is C(4,1) = 4!/(3! * 1!) = 4

So, picking 1 item in 4 will give you 4 options.

blueJanuary 25, 2012 at 6:15 pmthis is aint easy

jai prakashJanuary 29, 2012 at 4:46 amit is too good

cheyanne deniseFebruary 2, 2012 at 5:07 pmi luv this website it helps you out alot. thank god its on the internet for people that way they can have help on this kind of stuff. thank you for every thing.

kalidFebruary 3, 2012 at 12:45 am@cheyanne: Thank you! Really glad it helped out :).

Albert WignerFebruary 6, 2012 at 7:51 pmGreat discussion!

AnonymousFebruary 6, 2012 at 9:44 pmits all amazing….but not so fruitful

MarcFebruary 7, 2012 at 8:56 amI have a question. There’s a pool for soccer going around. 9 games every week. 3 options for each game. Win, lose or tie. How many different combinations are there in order to hit all 9 games right on 1 sheet and how would you even go about mapping that out?

AnonymousFebruary 7, 2012 at 5:18 pmThis was really helpful! Thanks a lot for the explanations!

kalidFebruary 9, 2012 at 10:30 pm@Anonymous: You’re welcome!

YahzidFebruary 8, 2012 at 2:12 pmThanks Kalid for your help but I am still having trouble understanding the concept.

I am still trying to understand the formula

There are 3 beats and a rest equaling 4 equal spaces of time all together.

|BBBR| BBRB|BRBB|RBBB|

I want to know How many different ways I can play 3 beats and a rest which equals 4 all together .

According to the definition this is a permutation because order matters.

The formula is C(4,1) = 4!/(3! * 1!) = 4

Now if the formula works I should be able to plug different rhythms and beat patterns into the formula and get all the possible patterns with no repetitions.

For example lets take 5 beats and 1 rest for a total of 6 events in time.

|BBBBBR|BBBBRB|BBBRBB|BBRBBB|BRBBBB|RBBBBB|

There are 6 possible patterns of 5 beats and a rest and now I will use the formula to see if it works.

The formula is C(6,1) = 6!/5!*1! = 720/120 = 6

Next example

Lets take 4 beats and rest for a total of 5 events in time.

|BBBBR|BBBRB|BBRBB|BRBBB|RBBBB|

The formula is C(5,1) = 5!/4!*1! = 120/24 = 5

Next lets take 2beats and 2 rest for a total of 4 events in time

|BBRR|BRRB|RRBB|RBBR|

There are 4 possible patterns of 2 beats and 2 rest

How do I put this into the formula ?

The formula is C(4,2) = 4!/2!*1! = 24/2 = 12

What am I doing or thinking wrong.

NakayamaFebruary 8, 2012 at 11:00 pmAh, finally i understood something. Arigato!

kalidFebruary 9, 2012 at 10:51 pm@Nakayama: More than welcome!

kalidFebruary 9, 2012 at 10:58 pm@Yahzid: Great question. In your last case (2 beats, 2 rest) the patterns are

|BBRR|BRRB|RRBB|RBBR| and also |BRBR|RBRB| for 6 in total.

To compute this, I’d take the total number of ways to arrange 4 items:

4!

and divide by the redundancies. There are 2! ways to re-arrange the Bs, and 2! ways to re-arrange the Rs. So we do

4! / (2! * 2!) = 24 / (2 * 2) = 24 / 4 = 6

One way to think about it: Imagine we label the Beats B1 and B2 and the Rs R1 and R2. We are dividing out the cases like this:

BBRR = |B1B2R1R2|B2B1R1R2|B1B2R2R1|B2B1R2R1|

As you can see, any pattern has 4 (aka 2 * 2) alternate ways to write it. The 4! formula shows every possibility and we have to divide out the redundancies we found.

Hope this helps!

ShabazzFebruary 14, 2012 at 5:03 pmVery halpful. why thank-you for this. i was struggling till i found this.

kalidFebruary 15, 2012 at 9:28 am@Shabazz: Glad it helped!

NickFebruary 15, 2012 at 8:15 pmYou’re the man. Very helpful. Better than my (foreign) teacher could explain it!

kalidFebruary 25, 2012 at 11:08 pm@Nick: Thanks!

MichaelFebruary 16, 2012 at 1:25 amHey, I noticed that you helped a lot of people with permutation and combination problems. I need your assistance.

I need to figure out this permutation/combination formula. How do I add the rule that objects within a particular set obey a hierarchy (i.e., the already predefined list of objects follows the rule that one object cannot appear below another object when it was originally listed above the object).

To better express my meaning clearly, here’s an example:

Set #1 – James, Jennifer, Jason, Amber

In the permuted/combinated list, James shall always appear above Jason and Amber. Amber shall always appear below Jennifer.

I want to add that rule, but I don’t know how.

Furthermore, I want to define this rule for more than one set. I want an equation that will permute/combinate different rules for different sets, but render all possible patterns based on those rules. Each set has 10 items, so the third set will contain only 10 items.

For simplicity sake, I’ll keep the number of items in the list down to six:

Ex. Set #1 – James, Jennifer, Jason, Amber, Florence, Carrie

Ex. Set #2 – Dustin, Michael, Morgan, Lance, Travis, Patrick

The rules are that names will be selected based on a preset pattern (i.e., names will be selected in a 4-6, 5-5, or 6-4 manner). For the first equation, each set maintains their respective hierarchy, but the hierarchy for set #2 does not apply to the hierarchy of set #1. Basically, 4 names will be randomly selected from set #1 and 6 names will be randomly from set #2, 5 names from both, and then 6 names and then 4 names. Once the formula/program solves the permutations/combinations, I want it to take those results and decipher the third set–in essence, match the results against the given but unknown information of the third set.

The third set will have numbers: 2, 2, 1, 3, 3, 4

Each individual in each set will have a rank designation.

Ex. Set #1 – James 2, Jennifer 3, Jason 4, Amber 1, Florence 1, Carrie 2

Ex. Set #2 – Dustin 1, Michael 3, Morgan 2, Lance 3, Travis 1, Patrick 2

Based on given information from the third set, the formula or program will most likely suggest a possible combination that suits the (2, 2, 1, 3, 3, 4) parameter:

James 2, Morgan 2, Amber 1, Michael 3, Jennifer 3, Jason 4

I’m no mathematician, but here’s how I visualize the equation in Microsoft Excel or something:

[(~set #1 & #2 – 4/6)(~set #1 & #2 – 5/5)(~set #1 & #2 – 6/4)] | (set #3)

** Please note that the selection process requires that both sets have 10 items (e.g., 4 items will be selected from set #1’s list of 10 items, 6 items will be selected from set #2’s list of 10 items). Under the aforementioned precepts, this “equation” represents all possible arrangements for these individuals. Then it matches those results and “deciphers” set #3.

I want to do this in excel or in another program, but I don’t know how to manipulate the permutation/combination formulas to obey new rules and so forth. I need help developing a workable formula.

kalidFebruary 25, 2012 at 11:12 pm@Michael: Hrm, interesting question. I think you need to structure it into groups: basically, define sets that can be re-arranged, and sets that can’t. If James must come before Jason/Jennifer, you have

James THEN Jason OR Jennifer THEN Amber

In this case there are only 2 possibilities (rearrange Jason/Jennifer). Each “OR” is a chance to use the combination formula to re-arrange the items in that little subgroup.

IgorFebruary 21, 2012 at 4:43 pmAn eaiser way is nPr where n is number of objects and r is number of items you want to select

MaiFebruary 29, 2012 at 7:48 pmI usually see the question asking for example: how many ways to arrange the letters in the word CHAIR? I know it is 5*4*3*2*1=120 but I have never seen the question asking how many ways to arrange the letters in the word ENTER, for example. Would you please explain it for me. Thank you very much.

CheraeFebruary 29, 2012 at 8:50 pmThanks! Helping my son with 6th grade math….wish you were his teacher!

or at least had written his textbook.

JulieFebruary 29, 2012 at 9:09 pmHi there!

I’m studying for a test and I was wondering if you could give me a hand with the following questions:

How many different 4 letter combinations can be made from the word SUCCEED?

I read your example earlier, where you treated it as if the doubles were different and then subtracted cases. But this time there are two sets of doubles which seems to complicate matters.

I thought I had finally found a solution by manipulating the combinations formula.

Since Combinations = (# of permutations ) / ( # permutations of objects picked)

I used the doubles formula for permutations ( 7P4 / 2! 2!) and then divided this by 4!.

But this got me a decimal answer. (8.75)

I tried it a few other ways too, usually getting answers between 10 and 30.

The textbook says the answer should be 230!!!!

I even tried listing the possible combinations, but can’t see it ever getting that high. Ack!

Also, if a question asks how many different amounts I can make from 3 quarters, 2 loonies and 4 toonies…..what is the easiest way to eliminate the cases where the two loonies will equal the same amount as a single toonie?

Any help would be greatly appreciated!!

Thank you so much!!!

CassieMarch 1, 2012 at 6:02 amHI

I have a question on coins and its really stressing me out.

Here it is:

A coin is tossed 8 times and the outcomes are recorded in a row.

How many of these have equal number of heads and tails?

According to my book, the answer is 70. I have been through calculations but none of them gave 70.

So would you help me on that? . Thank you very much.

tesnikMarch 6, 2012 at 9:41 pmHi Cassie,

Your book is right – the answer for the problem is 70.

Below is my explanation from an intuitive approach.

Understanding the quesiton:

————————–

As you’ve tossed the coin 8 times in a row, the sample space for the experiment will have 2^8 (= 256) outcomes. Your task is now to count the total number of outcomes that have equal heads and tails(ie.4 heads and 4 tails only) from the 256 outcomes.

Solution:

——–

To solve this, imagine 4 girls(~Heads), and 4 guys(~Tails) playing a simple game. You’ve 8 rooms in a row, and 1 room can hold 1 person only. You have to find out the total ‘combinations’ you can have with 4 Girls and 4 Guys. This will lead to the final answer.

To simplify it further, forget about the 4 guys, and just think of 4 Girls and 8 rooms (the remaining 4 vacant rooms will later be occupied by guys after girls choose 4 rooms for every combination).

So, 8 rooms and 4 girls leads us to –

8*7*6*5 = 1680, which is the permutation. But the order of 4 girls(heads) does not matter for the given problem, hence 1680/(4!) = 70.

Hope I did not confuse you :) . Email me if you have trouble understanding my explanation.

TesnikMarch 6, 2012 at 9:43 pmHi Cassie,

Your book is right – the answer for the problem is 70.

Below is my explanation from an intuitive approach.

Understanding the quesiton:

————————–

As you’ve tossed the coin 8 times in a row, the sample space for the experiment will have 2^8 (= 256) outcomes. Your task is now to count the total number of outcomes that have equal heads and tails(ie.4 heads and 4 tails only) from the 256 outcomes.

Solution:

——–

To solve this, imagine 4 girls(~Heads), and 4 guys(~Tails) playing a simple game. You’ve 8 rooms in a row, and 1 room can hold 1 person only. You have to find out the total ‘combinations’ you can have with 4 Girls and 4 Guys. This will lead to the final answer.

To simplify it further, forget about the 4 guys, and just think of 4 Girls and 8 rooms (the remaining 4 vacant rooms will later be occupied by guys after girls choose 4 rooms for every combination).

So, 8 rooms and 4 girls leads us to –

8*7*6*5 = 1680, which is the permutation. But the order of 4 girls(heads) does not matter for the given problem, hence 1680/(4!) = 70.

Hope I did not confuse you :). Email me if you have trouble understanding my explanation.

kaushik paulMarch 8, 2012 at 9:23 ami hav understood to to some dit.neway thankzzz

kaushik paulMarch 8, 2012 at 9:23 ami hav understood to some dit.neway thankzzz

zairMarch 9, 2012 at 4:04 amhi……

can u explain me with formula, how many combinations will occur?

thanks in advance.

(A,B,C) (E,F,G,H,I) (J,K,L,M,N)

markMarch 12, 2012 at 7:20 amhey,

i just love this site!! It reely helped me make a project… and my mom taught me this with the help of this site!! :) his site :)

thanks a lott… i hate maths, and permutations is a pain… i enjoyed it cozz of this site… :D

kalidMarch 12, 2012 at 10:59 am@mark: Awesome, glad you enjoyed it :).

GennisisMarch 12, 2012 at 6:18 pmI think I understand how to do it, but I am not sure which is the n and which is the r in word problems.

For example: In how many ways can a college registrar schedule four separate one-hour classes in a lecture hall during the four hours between 1:00 and 5:00 P.M.?

Moreover I am not sure if this is a permutation or a combination question. So maybe I really don’t understand…some clarification would be helpful.

Thanks ^.^

AmmarMarch 15, 2012 at 2:36 pmi have tow question and i need answer—>

1. Consider abox that contain 3 red,4 black and 2 white balls if tow balls are drawn from the box,one at a time with replacment,the probability that at least one of the tow balls is black

2. if three balls are drawn together from the box,the probability the exactly one of three balls is black is:

AnonymousMarch 17, 2012 at 6:47 amThis is awesome–simple and clear! :)

Muhammad ShahzaibMarch 20, 2012 at 12:36 pmThanks a lot…i could not pass in mid 1 because of confusion in determining the permutations and combinations

JasonMarch 20, 2012 at 8:03 pmThanks for the help, was having a hard time understanding this concept.

Bookmarked this site for future reference.

JerryMarch 23, 2012 at 5:54 amIm calculating the probability of a given number oft major earthquakes occur during a given period of time. Presuming the chance of a major quake is 151/1335 in any one motnth.

p= 151/1335

t= time no of months

n= no of quakes

Prob of t quaks occuring in n months if p for one month =

p^t * (1-p)^(t-n) * number permutations of n events in t periods

I currently do it programmatic with a binary count, counting the number of permutations having n ons. But as you can imagine for n>23 too long time

what is the formula approach

cu the number of ones

AnonymousMarch 23, 2012 at 7:01 amSorry for the double post, but my post came out very poorly from my mobile device. So here is it clean from my desktop.

I am calculating the probability of major earthquake occurrence rates.

over the last 112 years 151 major earthquakes occurred M7 or greater. That gives us an overall 151/1335 chance of occurrence in any given month.

p= 0.11310861423221 = 151/1335 chance of occurrence per month

t= number of months over which to find probability of occurrence of n number of occurrences.

n= number of occurrences

I am currently working with this.

probability of exactly n quakes (no more no fewer) occurring in t months if p is the overall probability for one month =

= p^t * (1-p)^(t-n) * the number of permutations of n occurrences in t times

Currently, I find the factor programmatically. I do a binary count of t bits and count the number of permutations which contain exactly n ones. because…

if a 1 represents a quake and 0 none then over t=4 months there are 3 permutations in which 2 quakes occur.

000

001

010

011 <

100

101 <

110 23 the count takes too long.

Is there a formulaic method for finding that factor? Seems the binary count might be a separate combinations problem. The result of which can be plugged into the quake occurrence formula.

Thank you and sorry for the double post.

Jerry

Relatively Interesting – The Birthday Paradox - Relatively InterestingMarch 27, 2012 at 5:35 pm[…] highly recommend the tutorial at Better Explained to brush up on the mathematics behind permutations and combinations – I […]

vivinMarch 30, 2012 at 5:37 amHi Kalid, i tried explaining permutation and combination to my sis, and i could succeed only in confusing her. i’ll try ur approach today. she has exams next week . Hope this works :)

Thanks !

AnonymousMarch 30, 2012 at 2:43 pm:)

AnonymousMarch 31, 2012 at 1:25 amhi I like this site very much ..

I have one Question ..

how could we find that , the given problem is of permutation or combination ? is there any ‘keyword’ ?

thanks…

AnonymousMarch 31, 2012 at 6:37 amhi,

can any one help me with the following problem please-

what is the total number of way N numbers can be picked such that the sum of the numbers is equal to S where the numbers can be from 0 to (s/2) ,repeatable and sequence dependent.

Thanks for any help.

rikMarch 31, 2012 at 6:41 amHi,

sorry if posting twice but I really need help in finding this

How many ways N numbers can be picked such that there sum is equal to S where the numbers can be from 0 to (S/2),numbers are repeatable and order dependent.

thanks for any help

jyothsnaApril 7, 2012 at 9:55 pmthanks a lot!! helped me understand the subject better…

muhammad Abul kalam azadApril 10, 2012 at 12:34 amExcellent lectures by which everybody can understand.

AnonymousApril 10, 2012 at 6:36 pmI LOVE U!!!!!!!!!!!!!!!

SamApril 11, 2012 at 10:31 amIn your combinations example Im still not understanding what happens to the 7. if your not actually multiplying it by the 3 then whats the point of having it there at all?

kalidApril 11, 2012 at 4:19 pm@Sam: Good question. Basically, the 7 is a placeholder telling us where to “stop.

If we have 10 items there are 10! (10 * 9 * 8 * 7 * 6 * 5 * 4 * 3 * 2 * 1) ways to order them. We don’t want to order all 10 though, just 3: so we pick the first 3 items:

10 * 9 * 8

But, we don’t have an easy way (on calculators) to say “Do 10 times 9 times 8 and so on, but only pick the first 3”. So we have to use 10 factorial to get the whole amount, and divide by 7 factorial to cancel out the items we want

10! / 7! = 10 * 9 * 8 [because 7 * 6 * 5 * 4 * 3 * 2 * 1 cancels from top and bottom]

You’re right that the “7” is more of a way to compute this easily with calculators. If you were figuring out possibilities on your own, I’d just do 10 * 9 * 8 directly.

mariaApril 13, 2012 at 8:29 amthank you soooo much, even though my teacher has been teaching for 32 years, but he couldn’t help me understand any word. I just read your lectures once, and understood everything !!!

kalidApril 13, 2012 at 10:11 am@maria: Thanks, really glad it helped! =)

JennyApril 16, 2012 at 7:01 pmDo you have any other info like this but with & without replacements, rules of sum & so on? Thanks!

GraceApril 18, 2012 at 6:20 pmThank you so much! I was seriously stuck on a problem until I saw this and now I actually get it! :D

carlApril 22, 2012 at 5:34 pmim confused can you show me how many permutations are in 3 letters and 3 numbers??

IanApril 28, 2012 at 2:21 pmYour article was helpful, however I find one problem on my homework to be extremely challenging for me- maybe I’m blind to something obvious:

At a neighbothood pizza shop, there are 5 veggie and three meat toppings. How manypossible pizzas can you order with one meat topping and one veggie topping?

Thank you very much for your help.

kalidApril 28, 2012 at 11:21 pm@Ian: No problem — break the problem down. Try writing it out, meats are Pepperoni, Sausage, Hamburger. Veggies are Onions, Mushrooms, Peppers, Cucumbers, Olives.

1) How many pizzas can you order with just one meat topping? (Ignore veggies for now). Someone says “I want a pizza with exactly one meat topping”. How many choices do they have?

2) How many pizzas can you order with just one meat topping and Onions or Mushrooms? It should be double of 1), since you have the Onion-version and Mushroom-version of each meat topping.

3) How many pizzas can you order with one meat topping, and any of Onions, Mushrooms, Peppers, Cucumbers, Olives? You should be able to write out the variations.

Hope this helps.

Evelyn Ofori-AmanfoApril 30, 2012 at 4:48 ampls. help me solve this question. As a transport manager ot the TOR, you have to plan routes for your drivers. there are six deliveries to be made to customers shell, allied, esso, bosch and turrow. how many routes can be followed?

swapnikaApril 30, 2012 at 10:16 pmdam gud explanation……wow..

vivMay 1, 2012 at 10:44 amplease answer a ques ….8 students qualify for final. all but 2 will advance to finals and out of those only 3 will get mealdals. findout how many such groups of 3 are possible.

Alexander Kruel · Number of combinations with repetitionMay 6, 2012 at 10:42 am[…] Easy Permutations and Combinations […]

ohthatcrazyblondeMay 6, 2012 at 3:49 pmThnx for the help! I desperately needed this!

naresh shahMay 8, 2012 at 7:31 amthere are 9 teams playing criket match aginst each other twice so what will be number of matches will be played and only top 4 team will qualify to play semifinal and final ?so,how many minimum matches the top 4 team has to win to qualify for semi and final.

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maishaMay 10, 2012 at 2:51 pmthis website really is going to help me on my test tomorrow. i have more confidence!

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mayaMay 13, 2012 at 12:12 pmthis is a lot of help, thank you so much:)

but i still don’t understand what k means

MinaMay 14, 2012 at 7:02 pmOMG! I actually learned it! Thank you so much!

HrishikeshMay 14, 2012 at 11:17 pmThank u very much. u’ve been a great help!!!

Maharshi DubasiMay 16, 2012 at 5:30 amThanx. That helped a lot. I am not joking.

SamMay 16, 2012 at 6:20 pmThis just saved me. Your explanations are so soooooooo helpful. I have a quiz tomorrow and I am a lot more confident now. THANK YOU!!!!

joshMay 19, 2012 at 3:19 pmyou must be a genus or something.are u a professor of mathematics or what?

kalidMay 19, 2012 at 4:21 pm@josh: None of the above, just a guy who loves math :).

yongMay 26, 2012 at 10:15 pmvery good

arishaMay 31, 2012 at 4:11 ami should combination and permutation examples

NicoleJune 3, 2012 at 2:29 pmThis helped me so much right before a big test!

SmritiJune 6, 2012 at 4:52 amHi

Amzing site. But I am still stuck with some questions .Please help asap

Thanks :)

If n people are seated in a random manner in a row of n theatre saets, what is the prob. That two particular people A and B will be seated next to each other?

If k people are seated in a random manner in a row, containg n seats, (n>k), what is the probability that the people will occupy k adjacent seats in a row ?

If k people are seated in a random manner in a circle, containing n chairs, (n>k) what is the probability that the people will occupy k adjacent chairs in the circle?

If n ppl are seated in a random manner containg 2n seats, what is the probability that no two people will occupy adjacent seats?

alodiaJune 16, 2012 at 3:31 amHello to the guy who loves math! Can you help me with these problems? Its some examples of permutations..

Three Boys & two girls are to be photographed. In how many ways can they be arranged in a row if,

a. if they are arranged alternatively

b.if no boy sits next to girl

c. if the row begins and end with a boy

AnonymousJune 28, 2012 at 9:20 pmHi,

Please help me solve this…

How many 5 digit numbers can be formed with 1-9 with exactly 4 different digits…

Navigate a Grid Using Combinations And Permutations | BetterExplainedJuly 2, 2012 at 12:14 pm[…] the ubiquitous combination/permutation problem — never thought it'd be useful, […]

savindaJuly 4, 2012 at 2:12 amhi friend’

please help me with this,

how many arrangements of 10 letters can be made from the letters of the word ‘PHILOSOPHY’. or words like that with repeated letters.

thank you

MimiJuly 4, 2012 at 7:45 pmThis definitely did help a lot. Thank you so much ^_^

rajeshJuly 5, 2012 at 12:54 amcan any body help me to solve the below problem…?

find the number of permutations of all letters of the word BASEBALL if the words are to begin and end with a vowel.

rajeshJuly 5, 2012 at 12:55 amcan any body help me to solve the below problem…?

find the number of permutations of all letters of the word BASEBALL if the words are to begin and end with a vowel.

rajeshJuly 11, 2012 at 8:06 pmThanks for your tutorial..it’s very easy to understand..

HimanshuJuly 19, 2012 at 6:36 amU should definately write a book with such a intresting method of teaching..it will help students to develop a interest in maths..

kalidJuly 24, 2012 at 4:50 pm@Himanshu: Thanks, I hope to write more in the future :).

docleoJuly 24, 2012 at 10:22 amI have four medications each with four different dosages. I want to mix three medications and also four medications using all possible dosages. How many permutations will I have using three medications and how many will I have using four medications? Can someone help me with this please?

HimanshuJuly 27, 2012 at 12:21 pmThats great sir…i am preparing for iit jee exams..i think you must be knowing about that..biggest entrance exam for engeenering in india…

Found this article really helpful as combinatorics is my weakest section in maths:):):)..

Wanna stay in touch with you…thanx

marrr14August 5, 2012 at 12:16 amhi can you share some application problems about these including the fundamental technique and compound probability problems, that would really help :) oh and with the answers at the bottom or something like that thanks :)

Diksha PandeyAugust 8, 2012 at 10:53 amIt was really a fabulous explanation 4 Permutation & Combination.Now I’m feeling much relaxed after deliberately going through this…..

Common Widget Elements | MuzzOctober 15, 2012 at 7:39 am[…] label) from all over the Internet: on a site homepage, in sidebars, in RSS feeds, etc. [reference: combination versus permutation] This widget is a combination […]

Permutations vs. combinations | For the love of spreadsheetsOctober 17, 2012 at 11:06 am[…] easy way to remember the difference between permutations and combinations, two terms which are often used interchangeably but incorrectly: Don’t memorize the formulas, […]

AnonymusOctober 20, 2012 at 2:21 amHiya! I would be really glad if you solved this problem.

Q. A ‘hand’ of 5 cards is dealt from an ordinary pack of 52 playing cards. Show that there are nearly 2.6 million distinct hands and that, of these, 575 757 contain no card from the heart suit.

On three successive occasions a card player is dealt a hand containing no heart. What is the probability of this happening? What conclusion might the player justifiably reach?

Please answer my question as soon as possible. Thanks..

How many bets are there?October 28, 2012 at 8:52 pm[…] from the total of N horses. For your other problems you may want to read a bit more about this: Easy Permutations and Combinations | BetterExplained Follow Math Help Forum on Facebook and Google+ « […]

minshiyaOctober 31, 2012 at 5:03 amwow! its a superb site……….

Si83November 4, 2012 at 3:00 pmHi, I’ve recently been taking an interest in combinations and permutations

As I’ve recently taken a job in a bookies and was asked the other day a question that I just could not figure out so I wondered if anyone here could answer it for me.

Q: If I picked 3 horses to win in 20 races, how many accumulators of 20 would it create?

Bare in mind that it’s 1 of 3 selections in all 20 races.

Can anyone help me figure that one please as its far beyond my capabilities.

shayanNovember 25, 2012 at 3:07 amHello ! i appreciate your work . Kindly answer and explain this :

You have 200 cards out of which 100 are Male, 100 are Female . What is probability of 2nd Female card before 3rd Male card ?

Dental Implants Santa RosaNovember 28, 2012 at 12:05 amGood post. I am experiencing some of these issues as well.

.

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AnonymousDecember 21, 2012 at 9:47 pmexcellent explaination. thanks a lot.

abhishek patilDecember 21, 2012 at 9:49 pmmaynnnn thax a lot its very cool and easy .

ShortcutDecember 27, 2012 at 3:05 amPls i nid a help

In how many ways may twelve persons be divided into three groups of 2,4 and 6 persons

the answer is 13860 but how did we get dat..

RohithJanuary 2, 2013 at 3:43 amLiked it a lot ,the way you arrived at the formula…Great.

Robert R.January 5, 2013 at 2:07 pmHow come you’re the only site that explains why this formula works? Thank you!

ChloeJanuary 7, 2013 at 2:05 pmEveryone loves it when people get together and share views.

Great blog, stick with it!

SHOBHANA suaraJanuary 8, 2013 at 11:12 pmCan anyone answer this ques:

Find the no. of ways u can fill 3×3 grid(with 4 corners defined a,b,c,d) if u have 3 white marbles and 6 black marbles??

1. 9C3 2.6C3 3.(9C3+6C3) 4.(9C3+ 6C3)/3!

Jason B. Hill – Courses » Class Meeting for January 28January 30, 2013 at 9:35 am[…] We finished the basic algebra review from Friday (see the link from Friday’s class) and also finished the discussion of permutations and combinations. […]

shifa masseyFebruary 7, 2013 at 1:57 amvery very nice :)

if i can understand; anybody can ..

GOD Bless yOu !!!!

kalidFebruary 11, 2013 at 11:25 amThanks Shifa!

shifa masseyFebruary 7, 2013 at 1:57 amvery very nice :)

if i can understand; anybody can ..

GOD Bless yOu !!!!

:)

annaFebruary 16, 2013 at 8:13 ampermutation and combination is quite confused i always forgt that

glumeFebruary 19, 2013 at 12:37 amFantastic site you have here but I was curious about if you

knew of any discussion boards that cover the same topics discussed here?

I’d really like to be a part of group where I can get responses from other knowledgeable individuals that share the same interest. If you have any recommendations, please let me know. Cheers!

LizFebruary 26, 2013 at 2:57 pmI have a question.

you are supposed to tell how many permutations and combinations can be made, im stuck on this question. (im only in 6th grade)

three of the letters A, B, C, D, E, F, and G

TraceyMarch 14, 2013 at 6:38 pmI have enjoyed reading your answers. You make a lot of sense and I appreciate it. However now I have a question if you can help me. In how many different ways can you select a committee of 3 people from a group of 13 memebers: The committee members consist of a chairperson, treasurer, and a secretary? I think it is a combination because it is not asking you to specifically pick who will be the c ,t, or s. It is just asking to select a committee. Is this correct? Please help.

kalidMarch 14, 2013 at 11:16 pmHi Tracey, great question. Actually, this is still a permutation because re-arranging the roles would matter (i.e., keeping the same people but switching roles would count as a different “group”). A better intuition might be “permutations are different if you re-arrange your choices, combinations are the same”.

So picking 3 people to *be in a group* is a combination, but picking *roles in the group* is a permutation. It’s really tricky, and I still mix it up. If it makes it easier, combinations work when every choice is “identical” (i.e., all the roles are the same), and permutations work when there’s something different between the roles.

So, that said… this is very similar to the “gold, silver, bronze” question above, but picking 3 from 13 (instead of 3 from 10).

Tracey RobinsonMarch 21, 2013 at 8:00 amHi Kalid,

Its me Tracey again with another question on combinations. Jaime is the Chairman of a committee. In how many ways can a committee of 5 be chosen from 10 people, given that Jaimie must be one of them?

would I do C 10!

4!(10-4) or do I keep the denominator as 5?

kalidMarch 21, 2013 at 8:16 amHi Tracey,

Very close. In this case, since we know Jaime must be on the committee, we’re really only picking 4 other members from the 9 remaining. So it becomes

9 * 8 * 7 * 6 / 4!

or

9! / (5! * 4!)

We divide 9 * 8 * 7 * 6 by the 4! re-arrangements within the sub-committee we’ve chosen.

Tracey RobinsonMarch 21, 2013 at 11:38 amThanks Kalid,

Your so helpful. I appreciat you taking the time to respond. Have a great day.

Tracey

anonymousMarch 21, 2013 at 1:52 pmI have 5 t-shirt design concepts I want to test (T-shirts A, B, C, D, E) but only want to show a random three to each person in sequential monadic format. I know how many combinations of three can be shown: (5 choose 3) = 10 But if I want to know how many times T-shirt B is in the 10 combinations, what is the mathematical way to do that. I know the answer is 6, but want to be able to figure out when I have more concepts. Thanks you for your help!

RaymondApril 1, 2013 at 12:18 amThanks a lot for the simple approach Kalid! I can understand more how it is like this now :) Thank you!

I think it would be even better if later on you could also explain how the other permutation rules work, like circular permutations, what if two people ought to sit together how do you arrange them, and the like :)) Wonderful work Kalid! :D

zainabApril 5, 2013 at 5:35 amthanks a lot yar. great

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Mohamad AmjadApril 14, 2013 at 11:47 ammy teacher kept trying to explain this thing for over two hours .. and everytime i looked to my classmate beside me .. we were like wtf with the crazy man ??!! and we just kept laughing out of confusion !!

YOU ARE A LIFE SAVER …

MaggieApril 15, 2013 at 4:32 pmMy teacher explained it to me, but i am still confused! If there are 8 sodas and you want to make 3 combinations out of it. How many combinations could there be?

LilyMay 5, 2013 at 3:06 amThis is my first time visit at here and i am actually pleassant to read all at one place.

jitendra khadkaMay 13, 2013 at 9:52 pmthis is also my first time visit at this website absoutely it help me ……….

SheldonMay 20, 2013 at 2:52 pmVery descriptive article, I loved that a lot.

Will there be a part 2?

SabraMay 21, 2013 at 12:29 amI can compute permutations and combinations all day long, but understanding how and why they work and, sometimes, when to use one over the other (only all the important parts, right?) can be confusing. The phrases ‘order does not matter’ and ‘order does matter’ in the definitions of combination and permutation confuse me. I think about the definitions differently, but want to make sure I am correct. Permutations=order matters=repetitions are okay and do count as new arrangements Combinations=order does not matter= repetitions are bad and do not count as new arrangements Though I KNOW this is faulty, the phrase ‘order does not matter’ computes in my brain as ‘order is not a factor in determining the number of arrangements’. With this thinking, ABC, BCA, and CBA would be considered 3 different arrangements. This, of course, is not what is meant by ‘order does not matter’ in the definition of combination because, in a combination, repetitions do NOT count as new arrangements. Now that I have explained the craziness of my brain, does my thinking about permutations and combinations in terms of repetitions counting as new arrangements or not counting as new arrangements seem valid? Thanks so much for taking the time to help.

KalidMay 21, 2013 at 2:18 amHi Sabra, great comment, thanks for sharing your think process. I agree, the language can be confusing/counter intuitive. Here’s another take: with permutations, shuffling counts as a new arrangement, with combinations, shuffling does not count.

It’s more important to find the wording that clicks with you vs. the standard “order matters” language which, I must admit, is not immediately clear. “Shuffles count as new items” or similar may click better!

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this post. I will probably be coming again to your blog for extra soon.

babylons ziomeJune 19, 2013 at 2:31 amthank you but its still a problem to me may you help me feather

saadiJune 30, 2013 at 1:47 ambuddy.thanx about me help.

sosanJune 30, 2013 at 11:13 amthanks…this site benifited me alot

AriJuly 1, 2013 at 4:56 pmJust wanted to say, this is an absolutely fantastic blog. Keep it up.

kalidJuly 5, 2013 at 9:49 amThanks Ari!

QroshJuly 13, 2013 at 9:08 amthaaaaaaaaanks ….

ShawlahJuly 30, 2013 at 11:45 amPLS,help me with this.

A woman has 11 close friends.

a)In how many ways can she invite 5of them to dinner?

b)In how many ways if two of the friends are married and will not attend separately?

c)In how many ways if 2 of them are not on speaking terms and will not attend together?

U’ve been of great help!!!

grapeJuly 31, 2013 at 12:49 amthis is great.. thanks a lot!

grapeJuly 31, 2013 at 12:55 ambut probbility is really confusing!! pls help :(

AnonymousAugust 4, 2013 at 11:02 amawesome

Ruben BarrosoAugust 5, 2013 at 9:59 pmThis post is great, but IMO in dealing with permutations you jumped to a “mechanical” explanation instead going further with the intuition path. Saying “We only want 8 * 7 * 6. How can we “stop” the factorial at 5?” and “And why did we use the number 5? Because it was left over after we picked 3 medals from 8. ” is not very intuitive to me. The way I would interpret it: “Hey, I initially have 8!=8x7x6x5! ways of giving 8 medals to 8 participants, where the order matters. This means that from every one of 8x7x6 way to give the first 3 medals to 3 participants, there are 5! ways of giving the remaining 5 medals to the remaining 5 participants (multiplication rule). But we are not really interested in how we distribute the medals among those remaining 5, so we focus on the first three medals, ending up with the initial 8x7x6 ways of distributing the first 3 medals. Of course, the mechanic translation of this is the formula you correctly state”.

Keep up the good work, Kalid!

kalidAugust 6, 2013 at 3:14 pmHi Ruben, that’s a great perspective/analogy, thanks for sharing!

DenaAugust 11, 2013 at 7:20 amwonderful issues altogether, you simply gained a new reader.

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ali bangash johns hopkinsAugust 23, 2013 at 12:19 amThey were simply holding their weekly service, prayer service,” Shea explained. She also fell into a fateful error- she married a cousin, Mahmud Muhammad Artan, and he took her virginity. Criminal convictions, civil judgments and financial irresponsibility will reflect negatively on the applicant’s character and fitness to practice law.

inderjitAugust 24, 2013 at 3:49 ami really very happy to find u r site…

harpreet kaurSeptember 1, 2013 at 4:52 amthanks.

ErnieSeptember 6, 2013 at 6:03 amGreat post.

LucySeptember 8, 2013 at 4:18 amFrom a group of 7 men and 6 women, 5 people are to be selected to be assigned to a committee in such a manner that at least 3 men are assigned to the committee.. How men ways are there to do so?? …. Please help.. Thank you!!

About 3d photographySeptember 8, 2013 at 8:06 pmIt’s very easy to find out any matter on net as compared to

textbooks, as I found this post at this website.

adam ramirezSeptember 23, 2013 at 11:39 pmC (44,6) how do i show work with these big numbers? i need to show my professor that i know how to work it out without typing it in to a calc…

adam ramirezSeptember 24, 2013 at 12:20 am4 golfers toss a tee to decide the order they will tee off (a tee is thrown in the air). whoever whoever the tee points to gets to go first. then the 3 remaining do it, and so on… what is the probability that they tee off in alphabetical order?

the answer is 1/24… if you could tell me why, it would be golden.

anveshSeptember 29, 2013 at 11:43 amthe num of 4 letter words that can b formed from the letters of the word “MEDITERRANEAN”

anveshSeptember 29, 2013 at 11:45 ampls help me with this question…….the num of 4 letter words that can b formed from the letters of the word “MEDITERRANEAN” that start wit E and ends with R is…. ans given 59

DenisonOctober 21, 2013 at 7:37 amThat “You tricked mE” is simply the best.. This was awesome..

JasiOctober 24, 2013 at 4:25 pmDo you know a place where I can practice the types of examples where you would arrange a group? For example:

PartA) A glee club has 10 members, and will send 5 entries to a duet competition. How many possible pairings are there?

Part B) Suppose that in each duet, one person is designated the lead singer, and the other will sing harmony. Now how many ways are there for the club to enter 5 duets in the competition?

One is multiplied and the other is divided by 5!

Why is that? please explain these types of problems.

HlayieOctober 26, 2013 at 7:05 pmIm writting my final exam on monday and i didnt get the concept well during lectures,and i search the net and found this,thanks Kalid your a real life saver.

kalidOctober 27, 2013 at 4:54 pmThanks Hlayie, glad it helped!

chrisOctober 29, 2013 at 1:41 pmThere are 3 sets of twins and 6 tickets. Two tickets are for row A, 2 Row B and 2 for Row C. Each twin selects a ticket randomly and sits in it that seat. What is the probability that each twin sits in the same row as his/her own twin? Dont even no where to begin…help!

FuckheDNovember 5, 2013 at 3:27 pmThis sucks

TimNovember 27, 2013 at 7:20 amHi There,

I have 8 football Games – I need to guess the results correctly, Home Win (HW), Draw (D) or Away Win (AW)

How many combinations are there so I cover every single outcome?

Hope you can asssit?

Best regards

Tim

KevinNovember 27, 2013 at 11:03 amI am stuck on this problem: Any suggestions?

A bit string is a sequence of digits consisting of only of 0s and 1s. How many 12-digit bit strings contain no consecutive 1s?

skamranJanuary 19, 2014 at 10:49 amI have been trying to solve the question in the post no. 41, “How many 4-letter combinations are there of the letters in each word? a) ONOWAY b) OSBORNE c) OUTLOOK”. I understand the logic of the first two but the third one(OUTLOOK) seems to have a different way to the solution. I know in order to eliminate the duplicates the formula should be C(6,3). But i don’t understand why we are counting other two Oz in it??

abhishekFebruary 8, 2014 at 7:46 pmHey kalid… Plz plz help me to solve this

AbhishekFebruary 8, 2014 at 9:01 pmAbhi n neetu park their cars in an empty parking lot with n>=2 consecutive parking spaces (I.e n spaces in a row where only one car fits in each space) abhi n neetu pick parking spaces at random.all pairs of parking spaces. Are equally likely.what is the probability that there is at most one empty parking spaces between them???

violetFebruary 10, 2014 at 7:38 amhey,. please help me with this question :

A witness told the police that the plat number contained the letters P,D,W followed by 3 digits, the first of which is 5. If the witness cannot recall the last 2 digits, but is certain that all 3 digits are different, find the number of car registration that the police may have to check.

Thank you so much :)

Esnart BandaFebruary 16, 2014 at 6:53 amits my first time on this site and i thank God for the oppotunity. been struggling with ths question for hours. please help me out:

how many nine digits numbers can be obtaind by using each of the nine digits exacty once?

how many of these are greater than five million?

ashwinFebruary 16, 2014 at 8:48 am@Esnart Banda

how many nine digits numbers can be obtaind by using each of the nine digits exacty once?

9 * 8 * 7 * 6 * 5 * 4 * 3 * 2 * 1 = 362880 numbers

how many of these are greater than five million(5,000,000) ( 7 digits)?

7! = 5040 numbers have 7 digits

362880 – 5040 = 357840 numbers are greater than 5 million.

I am not sure about the correctness of the answer…but this is what i could think of..do post the correct answer if you find it..

Ranjith BogodaFebruary 17, 2014 at 1:42 pmThanks lot for better explanation.

AnonymousFebruary 18, 2014 at 9:00 pmwasnt helpful

AnonymousFebruary 19, 2014 at 4:40 pmWhen can the number if permutations and combinations of n objects taken r at a time be equal?

wishwaFebruary 26, 2014 at 8:00 amhi sir, u r really outstanding,plz give some more basic solutions for the bignrs to pick it easily.

ericMarch 6, 2014 at 1:38 pma problem from biology: How many different possibilities are there to make a 100 amino acid long protein (there are 20 different amino acids, and they can be arranged in any order and can occur in any frequency). I assume it would be 20^100. This is neither a permutation nor combination. Is there a mathematical name for it?

Thanks

Great website, BTW!

kalidMarch 9, 2014 at 12:03 amThanks Eric! In a way, you can think of this as a sequence of 100 combinations: you have 20 choices, then 20 choices, then 20 choices, etc. So it’s really (20 choose 1) taken over and over again :).

linMarch 11, 2014 at 10:42 pmhow do i map a word problem into one involving permutation and combination?

e.g. if i have N slots to fill with a combination of size 1 and / or 2. Here, the order matters. e.g for N=5, 1+1+1+2 = 5 v/s 2+1+2 = 5 are different permutations.

HarithaMarch 12, 2014 at 4:00 pmThanks a lot …. I have a maths test just after 4 hours and and I finally know what this actually is !

teresaMarch 13, 2014 at 6:04 pmI don’t understand this problem could you help?

The instruction booklet for MM actually contains a chart showing how many possible codes there are, for any number of colors form 1 to 7, and allowing or disallowing repeated colors. Explain how you would go about making such a table. (You don’t have to calculate all of the values.) What will be the smallest an largest values in the table?

AnonymousMarch 21, 2014 at 3:40 amHi, my name is Monique:-)

This is really cool… Pretty much easier to understand.. Thanks alot… :-) :-) Awesome!!!!!

sheMarch 23, 2014 at 8:50 pmWhat a great explanation!

shahrozeMarch 25, 2014 at 2:54 pmFinally I got it….. Thankx a lot

KlohMarch 26, 2014 at 4:10 pmJust so things are clear for me,

Can I also say that of the 8! ways we can arrange 8 objects, (8-3)!

of them will start off with the same ordered triplet of objects

and the rest of the objects will arrange itself it (8-3)! different ways

and this is why we divide 8! with (8-3)! when finding the number of ordered

triplets in a list of 8 objects.

Is this a good way to explain it to myself or am I thinking about this wrong?

KlohMarch 26, 2014 at 4:13 pmI’d also like to mention that I am truly grateful for

all of this mathematical clarification you offer to the public

for no charge

Pyae Sone AungMarch 28, 2014 at 5:32 pmThank you very much, Sir.

AbarajithanMarch 30, 2014 at 4:38 amI’ve always loved ur site, I’ve recommended it to each of my friends. I love how u show something beautiful in math hidden from the textbooks, its like sherlock explaining his reasoning after a mess of confusion. Like sudden light in darkness, that aha moment, I love it.

However, this post is disappointing. Many textbooks offer this same explanation and its not helpful. Usually u get down to the level of the reader, assuming he cannot think math, and explain math in terms of common sense. U always write the article after getting urself a golden aha moment. Here, u didn’t do that. U assumed all readers are familiar with how the number of choices multiply (7x6x5), which is not the case. Or perhaps u urself has not felt that aha moment in permutation and combinations.

Sorry for my harsh words. I trust ur website so much and wanted this article badly. But its very disappointing.

melaniMarch 30, 2014 at 9:21 amthanks kalid. yo explanations helped me a lot to find out permutations easy.

thanks again.

JoshApril 1, 2014 at 2:26 pmSo brilliant! It really helps you giving a tutorials like this to people that learn differently. I am partially dyslexic and now I am understanding maths VERY well! I have become a kind of go to person in some respects for some problems in math thanks to you. :)

kalidApril 1, 2014 at 4:53 pm@Kloh: Glad you enjoyed it! Very close with a few corrections.

Of the 8! (40,320) ways to arrange 8 objects, there are 8!/(8-3)! = 8*7*6 = 336 starting triplets. For each triplet there are then 40,320/336 = 120 options afterwards. Note that 120 is 5!, or the ways to re-arrange the remaining elements.

@Pyae: Thanks!

@Abarajithan: Thanks for the feedback! Check out http://betterexplained.com/articles/how-to-understand-combinations-using-multiplication/ for more on why we multiply. As a quick check, imagine you have 3 shirts and 4 pants. How many outfits can you make? (Would it be 3 + 4 or 3 x 4?). Now, imagine you have 3 shirts, 4 pants, and 5 hats. How many outfits can you make? (3 + 4 + 5 or 3 x 4 x 5?). Hope that helps.

@melani, @josh: Thanks!

AlyApril 2, 2014 at 5:00 pmI really like your style!! The way u turned this scary topic into fun by magic :) u should definitely write a book. Cause your style is just what most kids need right now. I’m serious. No joking.

AnonymousApril 4, 2014 at 1:03 pmi Wonder how you became soo good???? i study but make mistakes at the examnation room

HELPApril 6, 2014 at 12:58 pmI AM SO CONFUSED

IsmailApril 8, 2014 at 11:44 amHow many arrangements can be formed of word”Equation” if all the vowels are kept together?

MahmudApril 11, 2014 at 2:19 amI get confused with the following question.

“9 different books are to be arranged on a bookshelf. 3 of these books were written by Charles Dickens. How many possible permutations are there if the books by Dickens are separated from each other?”

UmerApril 11, 2014 at 2:16 pmGreat site!

I have a question:

A choir consist of 13 sopranos,12 altos,6 tenors,7 basses.A group consisting of 10 sopranos, 9 altos,4 tenors and 4 basses is to be chosen from the choir.

1-In how many ways can the group be chosen?

2-In how many ways can the 10 chosen sopranos be arranged in a line if the 6 tallest stand next to each other?

3-The 4 tenors and 4 basses in the group stand in a single line with all the tenors next to each other and all the basses next to each other.How many possible arrangements are there if three of the tenors refuse to stnd next to any of the basses?

Desperately waiting for an explanation..I firmly believe that understanding the explanation of this question would help me understand this topic even more..!!

UmerApril 12, 2014 at 12:41 amAnswers to the above Question..:

1-33033000

2-86400

3-288

mwicigiApril 12, 2014 at 4:32 ami) In how many ways can an escort. of four solders be choosen from nine solders?

ii)In how many of these escort will a particular soldier be included?

chesterApril 12, 2014 at 5:34 pmreally better explained………….

apoorvareddyApril 13, 2014 at 6:51 amit was really a good experience

BrianApril 22, 2014 at 12:30 ami have two unequal lists of data, which are scattered through a list that is longer than either. So there are three possibilities A, B or nothing. How do I calculate the probability of A and B being on the same line ?

MikeApril 25, 2014 at 5:21 amAnother great explanation Kalid.

RachelApril 30, 2014 at 12:25 amThank you so much sir for such a lucid explanation.

This part is in our microbiology & as I didn’t have maths in 12th I was finding rather difficult to cope up with all this…

Please can you give me detail explanation with reference to binomial, Poisson’s and normal distribution… I’ll be very grateful to you sir.

Rahul YadavMay 5, 2014 at 10:49 amThis was very very very helpful. Finally, I get it.

WisdomMay 6, 2014 at 9:41 ampls sir, How many different circles can be drawn each of which passes through 3 of the 5 points a,b,c,d,e.if no.3 of the points are Collinear and no.4 are concyclic?…

permutations and combinations | ishizcoolMay 8, 2014 at 6:01 am[…] http://betterexplained.com/articles/easy-permutations-and-combinations/ […]

budhshivaMay 15, 2014 at 2:09 pmTHANKS A TON!!! I’m taking trigonometry at school, and I couldn’t even follow along during these past few lessons because I didn’t understand anything. Your website is extremely helpful :) THANK YOU!!

jacksonMay 19, 2014 at 3:41 amThere a four boys and four girls .In how many ways can they form a line, with the boys and girls alternating?

anonymousMay 19, 2014 at 5:35 pmVery helpful. Thank you for this site. I’m in seventh grade and even for me it is very clear. Thanks again

ArsalanMay 20, 2014 at 12:42 pm@Jackson: “B” will refer to boys and “G” to girls

As far as I can understand, this is a permutation. Hence order IS important. 2 types of orders are possible in this condition; Starting with a boys or starting with a girl

Consider a Boy/Girl situation.

Order: B G B G B G B G => 4 x 4 x 3 x 3 x 2 x 2 x 1 x 1 = 4! x 4!

Consider a Girl/Boy situation:

Order: G B G B G B G B => 4 x 4 x 3 x 3 x 2 x 2 x 1 x 1 = 4! x 4!

====> Total ways: (4! x 4!) + (4! x 4!) = 2(4! x 4!) = 1152 ways

This is based solely on my understanding. I’m a student myself and might have made mistake(s). Thanks :)

amosJune 14, 2014 at 2:08 amvery useful infact

SJune 23, 2014 at 7:28 amThank you, thank you, thank you!

I’ve been banging my head against this problem at Khan Academy for a week and was finally able to work through it. Understanding the “why” of things really helps.

AmitJune 24, 2014 at 7:17 ami have 5 positive signs and 3 negative signs i want arrange them in such way that the negative never come together?

Sengottuvel SJune 30, 2014 at 10:24 amThis is soo useful to understand at a very short time….thank u soo much….

HamzaJuly 3, 2014 at 12:28 amAfter practicing for hours and hours, I finally understood the difference. This was great help!

Pams OJuly 6, 2014 at 6:34 amHi. This site really helped me a lot. Thank you so much, Kalid. :)

BlairJuly 15, 2014 at 7:20 amGreat article. The explanation of combinations was brilliantly explained. Thanks.

kalidJuly 18, 2014 at 2:38 pm@Hamza, @Pams, @Blair: Thank you!

SrinathJuly 27, 2014 at 7:10 pmHi Kalid,

I have a question. I understand that the combination for getting 3 balls from 10 balls is 120. Could you please let me know the formula if the user is given 4 chances to pick 3 balls out of 10 balls.

Thanks

ManaliAugust 4, 2014 at 1:58 amif u pick a ball from 6red and 4 blue &

4 red and 5 blue what is the probability to get one red and another blue?

ASKED IN RBI GRADE B

GinnieAugust 31, 2014 at 6:51 amI have 7 tennis players, 38 weeks of tennis time, 4 players each week. Is it possible to schedule so that each player has equal number of days off and is scheduled to play with every player?

ChrisSeptember 5, 2014 at 7:12 amPlease explain the difference -which is a combination and which is a permutation of these two problems – I just don’t get it. Each assumes a standard 52 card deck.

1) How many different ways can you deal out 5 cards?

2) How many different 5 card hands exist?

anupamagcSeptember 9, 2014 at 3:58 am@Khalid: Could the explanation for #comment 30 also be given as:

No:of ways of selecting 1 question(6C1) + no:of ways of selecting 2 questions (6C2)+…+no:of ways of selecting all questions(6C6). I get the same answer (63 excluding 0) when I add them up

Also,why do we learn permutations first and then derive for combinations all the time, can’t it be taught the other way round?

jhilkeSeptember 11, 2014 at 9:32 pmfire brigade mangwale tu………angaro par hai aarma…….o balma o balma……….

LauraSeptember 20, 2014 at 8:57 amThis is incredibly helpful. I am studying for the GRE and, even with one graduate degree under my belt already, I have very little math background. Many of the quantitative reasoning questions are related to combinations and permutations and, although I had memorized the formulas, I was having a very difficult time applying them quickly and effectively, and I certainly wasn’t able to REASON well. This explanation helped me easily understand the reasoning behind the formulas and I am now able to quickly apply them where appropriate and solve problems much more efficiently. I particularly appreciated the use of humor in the delivery of these explanations. Humor is an excellent and underutilized learning and memory tool!!

AnonymousOctober 3, 2014 at 10:44 pmreally very useful…..thanks a lot

AnonymousOctober 3, 2014 at 10:45 pmthank u very useful…..thanks a lot

joshOctober 13, 2014 at 10:30 pmNice!

samOctober 22, 2014 at 7:05 pmPLEASE HELP

In how many different ways can you line up 6 students if students a and b must be separated by at leas 2 other students?

Muhammad FAizan ALiOctober 26, 2014 at 12:36 amCAlculate No of ways that 4 balls can be placed in 4 boxes

1.All are in different

2All are same.>

FlorinNovember 8, 2014 at 2:39 amHi!

If we have n heads, n bodies and n pair of feet, what is the formula to calculate how many different persons we can “build”?

hpNovember 14, 2014 at 8:20 amthis is damn good!

AnonymousNovember 19, 2014 at 7:48 amhi khalid,can u please give me the answer for this question,* piece of wood of length 10cm is to be divided into 3 pieces so that the length of each peice is a whole nuumber of cm,for example 2cm,3cm and 5cm-

a)list all the different sets of lengths which could be obtained

b)If one of these sets is selected at random,what is the probability that the lengths of the pieces could be the lengths of the sides of a triangle

draftedintomathNovember 23, 2014 at 10:34 amAn excellent explanation! I’m taking math on-line & it’s tough to get help, so this website was a lifesaver.

VNovember 25, 2014 at 12:46 pmHi…you have explained how 3 tins can be given to 8 people. Can you please explain how 8 tins be 3 people.

JazDecember 15, 2014 at 2:57 pmThank you! You explained this so much better than my professor!

Peter KelleyJanuary 12, 2015 at 8:07 amFirst: I am an Algebra impaired 55+ year old. I understand the theory of Algebra but I just can’t precess it consistently to be proficient. Secondly: I hope I have the right forum type to address my question. I think my question comes out of the combination/permutation discussion but it goes further in than a simple question of combinations or permutations. Here it goes!

Question: An imaginary planet has 4 moons. Each moon orbits the planet at a speed that permits EACH moon to occur with each phase of its other moons at key phases of:

Full Moon

First Gibbous / First Quarter

First Half Moon

Second Quarter

New Moon

Third Quarter

Second Half Moon

Second Gibbous / Fourth Quarter

[and back to New Moon]

I have tried with the timings of 8, 64. 512. and 4096 and find these do not cause the occurrences as I might have thought. I had thought that numbers ought to be divisible by 8 to permit the values of 100%, 75%, 50%, 25% and 0% to equate to precise phase representations. I can submit to you my spreadsheet and database review of my attempts at solving this off forum.

What would be the orbital times in days for these four moons?

Peter Kelley

St. Paul, MN USA

RbratJanuary 13, 2015 at 8:55 pmThis is awesome! My teacher didn’t explain the breakdown of (P) and (C) so, when she started talking about 2P1 and 2C1 I had no idea what was going behind the scene; however, now that you have clearly explained (P) and (C) it makes sense now. Thanks a lot. I am much more confident now that I have read your explanations.

Felicity AwayJanuary 21, 2015 at 4:44 amIt’s ridiculous how I understand this now when just this afternoon I was fumbling over notes and drumming trembling fingers on the desk of my chair scrambling for answers on the test. Thanks a lot! This is a totally helpful article, it just sounded so easy.

Oh no. You hear that?

It’s coming…

A-a…a stampede of-

:):):):):):):):):):):):):):):):):):):):):):):):)

AnonymousJanuary 22, 2015 at 2:11 amCool

monicaJanuary 22, 2015 at 7:21 amHey Kalid

How would you account for something like this:

16 exercises divided into 4 categories

you can choose 1 exercise from each category to create your own circuit workout

that is still a combination but it’s more limited because you can’t choose 4 exercises from just one category since those are in the same movement pattern family

thanks in advance for your help. I practically failed math in high school and college but reading this article helped me understand the basics of combinations so thank you for that!

kalidJanuary 22, 2015 at 10:08 amHi Monica, glad the article helped! In this case, it depends on whether the order of doing the circuit matters: is “pushups, jumping jacks, squats, lunges” the same as “lunges, squats, jumping jacks, pushups”?

If the order doesn’t matter, you can just multiply the choices available in each category:

4 choices in first category * 4 choices in second category * 4 choices in third category * 4 choices in fourth category = 256 choices

If the order does matter, we need to see how many ways we could re-arrange the categories. This is the number of permutations of 4 categories, or:

4 choices for category coming first * 3 choices for category coming second * 2 choices for category coming third * 1 choice for the last category = 24 category orders

So the total workouts, when category order matters, is 24 * 256 = 6144

Mark YujocoApril 27, 2016 at 9:43 amHi..!. A pleasant morning to you mr Kalid…… please help me about this problem….. 1. How many numbers in the range 1000-9999 have no repeated digits?

2. In a certain country, telephone numbers have 9 digits. The first two digits are the area code (03) and are the same within a given area. The last 7 digits are the local number and cannot begin with 0 . How many different telephone numbers are possible within a given area code in this country?

monicaJanuary 23, 2015 at 1:50 pmthank you so much for explaining. with some circuits order does matter so thank you for asking and for providing that answer as well.

MedaFebruary 8, 2015 at 2:13 amyou really are genius dude it seems like I understand the most incomprehendible topic in my life.

lalsingh sisodiaFebruary 9, 2015 at 3:22 amiwant to bob card how can applied in maater i have icici credited card but i want to bob card how can get its

kevinFebruary 15, 2015 at 5:22 amif 5 people enter a hall in which there are 10 vacant seats find in how many ways they can seat?

Falah IbrihemFebruary 18, 2015 at 10:39 pmHi , Can you give the answer of this questions please?

1-IF C (15,k)=C(15,2k-3) Find k?

2-IF P(n,r) = 120 C (n-r) Find r?

brad berryMarch 9, 2015 at 7:41 pmI’m trying to help my son with math and i am stuck…. with a total of $2.00 in coins how many different combinations are possible using a .50-cent piece, .25-cent piece and .10-cent piece ???? please help

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aditiMarch 20, 2015 at 9:20 amHi. I could not understand one point. You said choosing 3 people from a group of 10 is a combination but choosing President, VP and waterboy is permutation. Why is that? How does it matter who is selected first? How do we know which person of the three selected person gets which post? Please explain as this topic bothers me a lot. Thanks.

WaheedMarch 21, 2015 at 1:56 amHi Kaled,

Thank you very much for the intersting topics :) I am a big fan of your website.

I just have a question:

In European Champions League, There is a draw takes place in March every year to define what teams would play against each other among the qualified teams.

This year, we have 8 qualified teams:

Barcelona

Bayern Munich

Real Madrid

Monaco

Porto

Paris Saint Germain

Juventus

Atliteco Madrid

The draw will define the next 4 matches in this championship, The order is important here,

so “Barcelona versus Bayern Munich” (This match will take place in Barcelona) differs from “Bayern Munich versus Barcelona” (This match will take place in Munich)

The result of the draw would be like:

Barcelona versus Porto

Bayern Munich versus Monaco

Real Madrid versus Juventus

Atliteco Madrid versus Paris Saint Germain

The question is, how many possiblities can we have as a result for this draw?

My answer is (correct me if I am wrong):

P(8,2)*p(6,2)*p(4,2)*p(2,2) = 40320 (but this is really a huge number!!)

This is a video that explains how this draw goes:

http://www.youtube.com/watch?v=yOTgkp1D4yo

Thanks in Advance

Mike HMarch 23, 2015 at 11:11 amSo if I understand this correctly…

r-permutations are really the total permutations divided by the permutations of what it is not.

ex: P(10,3) ways to choose 1-3rd place out of 10 people

total perm = 10! = 3628800

perm of those we are not counting ( 7!) = 5040

total/perm of not counting = 3628800/5040 = 720.

and combinations are that number (720) divided by permutations of different ways to order 3 (3! = 6)…

thus 720/6 = 120 = C(10, 3).

I know this is convoluted but I’m trying to grasp this at fundamental level because I dont understand bit strings. How many bit strings of length 10 have four 1s. supposedly the answer is C(10,4) but I cant find an explanation of why that is. the 10 in that scenario I thought was reserved for the size of the set or the number of ways on thing could be chosen like C(52,5) for cards. _ _ _ _ _. the number of choices for 1st space is 52, then 51 etc. and 5 represents the number of combinations we are counting.

how does the fact that I only have 2 choices for each digit come into play ( in the bit string example). total number of permutations I thought would be 2^10 = 1024 and number of r-permutations (r = 4) would be …I think im actually losing myself here. Maybe I missed it as I was scrolling through comments, but could you explain combinations or permutations of bit strings. and perhaps how the choice of only 1 or 0 comes into play there? you can use the example of bit string length 10 choose 4 if you want or why i use C(10, 4). supposedly being (10!/6!)/4!= 210. Why is it 10*9*8… int the numerator. there arent 10 choices * 9 choices… im lost :) and I love your work by the way. I have already recommended your site to fellow students.

Kobi WessApril 8, 2015 at 10:07 amHELP!!!!

I am truly stuck on this question…..

To win at LOTTO in one state, one must correctly select 7 numbers from a collection of 47 numbers (1 through 47). The order in which the selection is made does not matter. How many different selections are possible?

I know to use the formula for a combination and I can determine the values of n and r….but i get stuck at simplifying to get the correct answer…..

JustinApril 9, 2015 at 5:25 amThanks.Now I got what really combination means.Could you explain why if you take 3 heads out of 5 coin tosses your value of k!=3!=6 while p(5,3)=60.I’m confused why there is only 6 variants since for each one of the 60 outcomes you could change it for 3 heads and 2 tails so it would be 60*3!*2!=720?I took heads as H1,H2,H3 and tails as T1,T2.To place heads in 3 positions there are 6 ways and to place tails in 2 positions there are 2 ways.So for 60 outcomes it would be 720 ways.Could you help me.

KARTHIK AMARNATH SAAKREApril 10, 2015 at 10:42 pmI always got confused with the Permutation and Combinations.

Your Explanation saved in my mind for permanent .

Thank You Sir..!

Erik SutugaApril 17, 2015 at 8:18 amit wasgreat explaining

IanApril 20, 2015 at 4:43 amConsidering a coin flipped three times and wanting 2 heads, in what sense doesn’t order matter that gives us 3 combinations? What are the six permutations?

BallyApril 28, 2015 at 5:09 amCan anyone tell me: If I have 4 tops, 4 jeans, 4 coats and 4 shoes, how many outfit combinations can this make? Thanks

ThandoMay 2, 2015 at 10:39 pmHi Kalid, this really helped me a lot! Thank you for that. But about the question how many different numbers above 6000 can be formed from the digits 3,4,5,6,7…

The question seems a little ambiguous to me. It doesn’t say the number has to have 4 digits, just that it has to be greater than 6000. Also, it doesn’t say whether or not the digits may be repeated. The way I understand this is that it is therefore any number less than infinity and greater than 6000, so long as it has those 5 digits. Please help!

karenMay 4, 2015 at 12:29 pmThanks for your great explanations. I understood the combinations but couldn’t find good explanations of the notation. For the 2nd question above (the one about pulling 2 face cards from a deck,) could you work it out completely? Also, for the 3rd question above (the one about forming numbers larger than 6000,) wouldn’t you also have to add the five digit-numbers which you could make? Like 34, 567, etc.? Thank you!

JackMay 5, 2015 at 11:44 amI have 6 items – 1A, 1B, 2A, 2B, 3A, & 3B. I’m trying to develop a formula to determine the number of possibilities if 1A and 1B, 2A and 2B, and 3A and 3B cannot exist together. I need groups of 1 (the answer there is 6); groups of two (where 1A-2A is legal but 1A-1B is not– Should total 8) and finally three (1A-2A-3B is legal but 1A-1B-2A is not- Should total 8). I worked it out graphically but if the number of items is 20 instead of 6 it turns into a graphic nightmare. And to take is one step further lets add 4 and 5 (without A or B) and up to 5 possible items.

UpanshuMay 16, 2015 at 11:49 amthere are three different brands of pens blue, green and red. then, how many ways a person choose 4 pens from the above with repetitions allowed ?

Sameer ( Hindu Indian)May 17, 2015 at 6:47 amWow that was quite interesting and finally I understand it . Thanks a lot .

PriyaMay 18, 2015 at 7:40 amThis explanation is very understanding. Every one can get it very easily.

Thank you so much for better explanation.

Logan EnfingerMay 19, 2015 at 11:16 amI have a quiz in a few minutes on this. This helped greatly. Thanks!

milanMay 22, 2015 at 11:03 pmWOW..!! That was really a easy way to understand permutation and combination.. Thanks a lot.

lalalandMay 25, 2015 at 3:26 pmi love this this really helped me out cause i got these questions on my scantron

kkkkMay 27, 2015 at 6:24 amHi I m stuck on this question …can someone help.

for some reason the net is not coming with lines but u can make it out

Question :A set of children’s blocks are to be painted in distinct colour patterns. The blocks are made of cubes and must be painted so that no block is able to be rotated into a position that matches the paint scheme of another block. Each face must be painted a unique color from a set of 6 fixed colour choices. A net showing one such colour scheme from the set is given in Figure 2.

2

3 4 5 6

1

1 – Red

2 – Blue

3 – Green

4 – Yellow

5 – Pink

6 – Orange

Figure 2: Net representation of a cube and colour pattern.

(a) If face 1 is painted Red, how many ways could the opposite face (face 2) be painted?

(b) If face 1 and face 2 are painted, how many ways can the remaining four faces be painted?

(Hint: because the cubes must be painted in distinct patterns you will need to ignore the four possible rotations which could result in a non-distinct pattern, e.g. {3,4,5,6} is the same as {4,5,6,3}, {5,6,3,4} and {6,3,4,5}.)

(c) What are the total number of ways the set of cubes can be painted?

(d) If 7 fixed colour choices were available instead of 6, how many distinct ways could the blocks be painted?

kkkkMay 27, 2015 at 6:30 am2

3 4 5 6

1

hi this is the correct ne 2,4 1 need to be vertically in line

thnx

Saumil JoshiMay 31, 2015 at 2:23 amCould you explain Diagonal sum’s in pascal’s triangle intuitively and permutations and combinations with repititions ?

mokeeMay 31, 2015 at 11:47 amit is very nice lecture method, please continue such like that more!!!!!!!!!!!!!!

AngeloJune 1, 2015 at 7:55 pmhow many combinations of three different numbers can be chosen from the numbers 23 through 32? I don’t understand why this is a combination. It says three DIFFERENT numbers so that means you can’t have 23 23 23 right? Doesn’t that make it a permutation?

Puneet RaiJune 3, 2015 at 11:47 pmplease explain how to find the rank of a word

for eg. ZENITH

clintonJune 4, 2015 at 12:25 amthank kalid for that

JJJune 8, 2015 at 10:20 amHi Kalid,

Your explanation was somewhat like my teachers’ but I wanted to ask if you can put more practice problems with permutations and combinations on your site and apply it with an answer key so that beginners like me can practice. I am still having trouble with permutations and combinations though. Thanks for your explanation anyways and I hope you can post practice problems.

noor hasinahJune 9, 2015 at 6:49 amcan someone help me please…

Malini decides to select 5 songs from a list of 10 songs to be sung during a concert. in how many ways can she

a) make a selection

b) presents the songs

AnonymousJune 9, 2015 at 3:18 pmThanks for this site. My test is tomorrow and this helped me understand the study guide SOO much better! THANKS!!!!!!❤️

HipersonJune 9, 2015 at 3:27 pmThanks for this site. My test is tomorrow and this helped me understand the study guide SOO much better! THANKS!!!!!!❤️!

HyperpersonJune 9, 2015 at 5:04 pmThanks for this site. My test is tomorrow and this helped me understand the study guide SOO much better! THANKS!!!!!!❤️!:)

HyperpersonJune 9, 2015 at 5:07 pmHyperperson says:

Thanks for this site. My test is tomorrow and this helped me understand the study guide SOO much better! THANKS!!!!!!❤️!:)

BigGnomeJune 10, 2015 at 3:52 pmThank you, my Final Exam is tomorrow and this was a big help.

math hates meJune 10, 2015 at 10:15 pmthnks a bunch

AdityaJune 30, 2015 at 12:39 amThank you!

rahulJuly 19, 2015 at 12:21 pmAwesome Explanation.. finally I am able to learn what permutation and combination is exactly.

Soban KhalidJuly 23, 2015 at 1:54 ambro can you re explain the second question from ” Questions and contributions ” I couldnt understand it actually :/

AndrewJuly 28, 2015 at 3:21 amI love your site and would like to practice with more examples. Any recommended sites

Smruti SwadhismitaJuly 28, 2015 at 8:46 amMaths becomes when u understand the theory..

I want to learn more..

Sadiq RazaJuly 28, 2015 at 11:24 amThanks a lot…!!! it clears the confusion.

AnonymousAugust 11, 2015 at 5:02 amHello, thank you very much I have understo od your explanation…

rishAugust 12, 2015 at 11:45 pmi love u

raisuAugust 13, 2015 at 9:23 amits very nice…………..but……i want …..clear answer for 31 comment.help me ..due…

AnonymousAugust 14, 2015 at 5:34 pmPerm mutation does sound like a relay hairy problem !

ScottAugust 24, 2015 at 3:23 amI am really having problems inputting this in excel. let’s say I have 15 friends and 3 presents to hand out to them. How many different ways can I hand these presents out using permutation. How could I input this as a formula in excel and show the outcomes to my daughter so she understands more. Thanks.

BuboyAugust 27, 2015 at 7:29 amWhoa! Thanks a lot! It makes clear sense now :D

shaliniAugust 29, 2015 at 6:10 amit is very easy to understand from the internet that all the maths problems very clearly.

eric amanzeAugust 31, 2015 at 1:18 pmI am predicting results of 3 football matches (A,B,C) where there are only two possible outcomes, a draw or a win.

My question is, how 3 match outcomes are possible giving these indices?

if there are 4 matches instead (ABCD) how many 4 match outcomes are possible?

eric amanzeAugust 31, 2015 at 1:26 pmI am predicting results of 3 football matches (A,B,C) where there are only two possible outcomes, a draw or a win.

My question is, how many 3 match outcomes are possible giving these indices?

if there are 4 matches instead (ABCD) how many 4 match outcomes are possible?

Ian CabSeptember 2, 2015 at 5:51 am1) In how many ways can 3 dogs, 4 cats, 4 rabbits and 2 hedgehogs be seated in a row?

2) How many different signals, each consisting of 7 flags hung in a vertical line, can be formed from 4 identical red flags and 3 identical blue flags?

3) A classroom has 10 vacant seats. If five additional students enter the class, in how many different ways can they be seated?

4) In how many ways can five students be seated i a row of ten seats if two students refuse to sit next to each other?

5) In how many ways can 15 students take 3 different tests if 5 students are to take each test?

Those items in my homework kinda confuses me. Help me pls Thank you so muchhhh!

ToriSeptember 5, 2015 at 6:50 pmHi, I have a problem and don’t understand it and was hoping you could help. Here’s the problem. -6+19C5

I tried solving it but didn’t understand it.

AB Rashid DarSeptember 8, 2015 at 2:11 amvery useful sir……sir i need full help in understanding the concepts n practical examples of probability

RenSeptember 12, 2015 at 8:12 amThe letters of the word BRONZE are written in all possible orders. if these words are written out in a dictionary order what is the rank of the word BRONZE ?

AnonymousSeptember 16, 2015 at 3:21 amThanks mate. Real big help.

KONIGSeptember 17, 2015 at 6:04 amThis is awesome, thanks dude

FarhanSeptember 17, 2015 at 10:01 amHello , please help me with this:

1 – A bag contains 5 red balls and 7 green balls. find the number of ways 3 balls can be drawn from the bag if:

(a) 2 must be red and 1 must be green

(b) they must all be of same colours

2 – a committee is to be formed containing of 7 members from 8 parents , 5 teachers and the principle.

how many ways can the committee be formed if the committee consists of;

(a) the principle

(b) 4 parents

KishanSeptember 21, 2015 at 11:53 amA very good explanation here. I was always confused in selecting either permutation or combination. Now its crystal clear thanks to you.

SandeepSeptember 22, 2015 at 7:13 amThank u ……. I had big time trouble before I saw this site…..u r a savior!!! :)

NobodySeptember 23, 2015 at 6:08 amHow would I calculate something with much bigger numbers? Say, for instance, a protein that is 120 amino acids long, with 20 variants of amino acid choice? Would I use the same formula? Thanks!

TundeOctober 5, 2015 at 12:56 am@Kalid

You sure are doing a great work here making things that look really hard/challenging/tasking quite easy.

Thanks

AshwiniOctober 6, 2015 at 10:52 amFrom 0 to 9, how many permutation combinations are there?

Explain me this question with solution

AsulMarch 30, 2016 at 5:17 am1

heheApril 5, 2016 at 7:01 pm2

Fernando EnriquezOctober 12, 2015 at 7:08 pmI have a question and I cant figure it out. its off of my daughters homework and its driving us crazy.

A cereal company is giving away baseball erasers free in their boxes of cereal. there are 7 erasers to collect. all the children at the school want the whole set. at any moment, they all have different collections and none have more then one of the same eraser. (The collection can overlap – for example, one child could have a,b, and d while the other has b,c, and d)

Ali Sulaiman MohammadApril 1, 2016 at 4:32 amI should think. 7C0+7C1+7C2+7C3+7C4+7C5+7C6+7C7

10th grade student

Fernando EnriquezOctober 12, 2015 at 7:12 pmquestion 1. what is the maximum number of children there could be at the school?(things to think about: how many children could there be if there were 2 erasers to collect? if there were 3 to collect and so on. can you find the pattern) remember, it is possible for one student to have no erasers.

marc ace tanOctober 14, 2015 at 8:38 pmfrom the first 4 odd numbers (1,3,5 and 7) and the following 4 even numbers (2,46, and 8), how many groups of 4 digit numbers can be formed consisting of:

a. 2 odd numbers and 2 even numbers?

b. all odd numbers?

c. 1 odd number and 3 even numbers?

Waizah ChanOctober 22, 2015 at 6:30 pmVery awesome and helpful discussions… Thank you ALL!!!

IdinaOctober 25, 2015 at 7:31 amWhat are the best resources u can recommend to learn Probability very effectively?

Thanx in advance!

VimalkumarOctober 29, 2015 at 12:23 amI have five colors I have to match it as three combos…can anyone provide me the exact combinations count ..

K.Aditya SaiApril 13, 2016 at 6:21 amC (5,3) =10

steveOctober 31, 2015 at 1:54 amSuppose you have 13 football matches and each game has 3 chances win, lose & Draw. How can you get a chances of all possible outcomes?

rofemieNovember 3, 2015 at 5:06 pmcan you help me with this one?

if there are 3 roads from town a to town b and 4 roads from town b to town c, in how many ways can one go from town a to town c and back and to town a, through town b without passing through the same road twice ?

vincent nzauNovember 4, 2015 at 1:47 amsuppose there are 13 bags each with three balls;red, green and blue, and you pick a ball from each bag. how many combinations are available? can also get the list of the combinations?

vincent nzauNovember 4, 2015 at 1:52 amsuppose there are 13 bags numbered 1-13, each with three balls;red, green and blue, and you pick a ball from each bag. how many combinations are available? can also get the list of the combinations, in order of the bag number?

PeterGNovember 5, 2015 at 8:10 amHi Kalid,

Very wel explained indeed. I am no math wiz but I have this issue: I have 27 different items, each item having a value of Yes or No. How many possible combinations will I have? Thanks for the help. :)

P.S.

I am trying to put the combinations in Excel and I am wondering if I can display all the combinations in Excel, i.e., if Excel has enough columns to display all the combinations.

AnonymousHelperApril 13, 2016 at 4:56 amAnswer: 351 (from 27C2)

where:

–> 27 = number of items

–> C= combinations

–> 2 choices (Yes or No)

Sorry for the late reply haha :-)

P.S. found out the answer using the calculator.

skNovember 10, 2015 at 6:36 amthis was awesome!!!!!!

DenNovember 11, 2015 at 5:13 amCan you give an example of real life that involve permutation at least 3. Cause I’m very confusing what permutation is. Reply pls ASAP

-Den

rarin2dinNovember 16, 2015 at 9:22 amHow do I solve this….

A password has 5 letters and can be made from the letters A,B, and C.Letters can be repeated.How many different passwords are possible?

CarlosMarch 26, 2016 at 3:04 pmThis is a permutation with repetition. therefore you have n^r permutations to choose from: 5^3.

rarin2dinNovember 21, 2016 at 12:48 pmCarlos,

You mean 3^5 = 243 . Since the password has 5 letters and each can have 3 choices.

AnonymousNovember 20, 2015 at 9:40 pmI do really like this site. But I have a question in my book that says:

A white wolf horse is in a race with 7 other runners. If we are concerned with 1st, 2nd and 3rd placings, in how many ways can white wolf finish 1st or 2nd or 3rd?

dollar treeNovember 23, 2015 at 6:06 pmThis site really helped a lot

SAMARKNovember 23, 2015 at 8:20 pmHI I NEED HELP HERE…..

How many ways can we award a 1st, 2nd and 3rd place prize among eight contestants? (Gold / Silver / Bronze)

8 choice* 7 choice* 6 choice.

My question is, why are we multiplying these terms? isnt this addition here? 8+7+6?.. please help, i need answer fast….

PrinceNovember 27, 2015 at 3:28 amA big round of applause to u for making this thread. You should be my teacher……. Thnx a lot…..

SUFYNovember 28, 2015 at 11:21 pmthank you… it was help full .. from school onwards I was little bit confused about permutation and combinations. and I always wonder how could I understand the question, whether it is combination or permutation. So I was skipping this portion and study all other portions for exam.. Now I got the idea and I am sad to know this is very simple.. thanks for helping me..

O. OparaDecember 1, 2015 at 1:02 pmDear Kalid,

I’ll say you are really a blessing to me. I so much value the way you make difficult things turn so easy. I am really so so happy for stumbling upon your site today after so much search for better explanations on different topics I find so confusing.

I’m one of the many that loves maths a lot but also finds it so frustrating most of the times, especially when the exercises are so twisted than what was used in explaining the topic.

I have two sets of exams to write on maths in the next 2 weeks and have been facing so much difficulties in some topics.

I believe I have reached a point where I can grab a sit and learn comfortable and as much as I can before my test.

Thanks a million bro!!!

kalidDecember 1, 2015 at 8:50 pm@O. Opara: Really glad it helped!

shizaMarch 28, 2016 at 12:01 pmthanks for this easy task…..

it helps me more clear me the things…

plzz give more examples or exercises for probability solving and giving experiences for us

thankssss

AnonymousApril 1, 2016 at 7:49 amAwesome explanation

KISHAN LALApril 8, 2016 at 1:30 amI HAVE UNDERSTOOD THE DIFFERENCE TODAY AFTER A GAP OF 38 YEARS

AKNApril 10, 2016 at 11:27 pmWell written!

ManishaApril 14, 2016 at 7:10 amvery well explained

ManishaApril 14, 2016 at 7:13 amI loved the way he has explained. superb

Ibikunu BrownApril 14, 2016 at 12:46 pmI love the way u’ve broken down. Thanks. However, I still have other challenges on integration, indices and logarithm. you can help same way or better still, drop it in my mail box.