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:

## 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.**

## Leave a Reply

1256 Comments on "Easy Permutations and Combinations"

[…] (Brush up on combinations and permuations if you like). […]

how to do 6c3

in your calculation, press 10 and press C (Combination) of and then press 3

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

it’s 6p3/3p3 or (6*5*4)/3*2*1) which = 20.

It is 6•5•4/3•2•=20

6*5*4*3*2*1/6-3!*3!=6*5*4*3*2*1/3*2*1*3*2*1*=720/36=20

can 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

easy suck my

Hello,

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.

Good explaination

Thanks alot! This was actally a better explanation then my teacher could give us =]

hallelujiah!

hala sagol

hahahahahahaha

hahahahahahaha

yea, same!

kooooty

a uper sara bakwas lika geya a muhalil nay sahe mae ny ma hansa a muhalil navy ma kam karta a

Agreed – this article helped me understand it WAY better.

Awesome! Glad you found it useful :)

finally this makes sense

This still makes 0 sense

I 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!! :)

that was very helpful thankyou

this is an awesome site!

tomar baal xD

Thanks a million! It makes sense now!

Thanks for the comments, glad you found it useful.

If my chances are 1 in 13 million of winning the lottery and I buy 10 tickets, do my chances increase?

yes they will be 10 in 13 million kinda obvious

Kinda 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

yes

Hi 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 :).

Im not really sure but I dont think Kalid cares after nearly 9 years

WTF

WTF 2

WTF 3 …. DOTA 3 Confirmed

wtf 9 years!!!

Did Cooper win the lottery after all?

I’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!

So what!! These are the number of ways .

Hi 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).

This helped with my question… sort of

Great Stuff! You should write a book!

Thanks for the encouragement Aaron! Once I have enough posts I would love to turn it into a book :)

it was really useful dude!!!

thnx a lot!!!

No problem, glad it helped!

Worst planning

I 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

if 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?

Thanks alot!!!!

am studying n i have an exam 2mmorow !!

thnx 4 helpn me

Hi Besho, I’m glad I was abble to help!

10 pairs of shoes are well mixed up.4 shoes are randomly picked. What is the probablity of getting at least 1 complete pair

10c1*18c2

This is really an excellent way of explaining the things!!!

Hi 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 :)

No. 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

hey

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

I 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..

Hi 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

hi!!! where can i find the solutions to the problems stated above???