Easy Permutations and Combinations

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).

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 items. 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 pick a Gold, Silver, and Bronze medal for “Best friend in the world”?

permuation_medals.png

We’re going to use permutations since the order we hand out these medals matter. 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: \displaystyle{8! = 8 \cdot 7 \cdot 6 \cdot 5 \cdot 4 \cdot 3 \cdot 2 \cdot 1 }

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:

\displaystyle{\frac{8!}{5!} = \frac{8 \cdot 7 \cdot 6 \cdot 5 \cdot 4 \cdot 3 \cdot 2 \cdot 1}{5 \cdot 4 \cdot 3 \cdot 2 \cdot 1} = 8 \cdot 7 \cdot 6}

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:

\displaystyle{\frac{8!}{(8-3)!}}

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:

\displaystyle{\frac{n!}{(n-k)!}} just means “Use the first k numbers of n!”

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

\displaystyle{P(n,k) = \frac{n!}{(n-k)!}}

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 going to be equally disappointed.

This raises and 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

\displaystyle{C(n,k) = \frac{P(n,k)}{k!}}

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:

\displaystyle{C(n,k) = \frac{n!}{(n-k)!k!}}

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 - it’s better to know why they work. Combinations sounds simpler than permutations, and they are. You have fewer combinations than permutations.


Posted February 12, 2007, under Math
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    Comments

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

      PJ — May 3, 2007 @ 7:33 pm

    2. Awesome! Glad you found it useful :)

      Kalid — May 3, 2007 @ 10:14 pm

    3. finally this makes sense

      Anonymous — May 10, 2007 @ 7:13 am

    4. this is an awesome site!

      Kenny — May 22, 2007 @ 4:08 pm

    5. Thanks a million! It makes sense now!

      Sheri — May 25, 2007 @ 10:31 am

    6. Thanks for the comments, glad you found it useful.

      Kalid — May 25, 2007 @ 11:53 am

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

      D Cooper — June 5, 2007 @ 6:18 am

    8. 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 :) .

      Kalid — June 5, 2007 @ 10:40 am

    9. 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!

      R David — July 17, 2007 @ 7:03 am

    10. 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).

      Kalid — July 19, 2007 @ 2:29 am

    11. Great Stuff! You should write a book!

      Aaron Chaon — July 30, 2007 @ 4:42 pm

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

      Kalid — July 31, 2007 @ 3:15 pm

    13. it was really useful dude!!!
      thnx a lot!!!

      Dheeraj — August 1, 2007 @ 3:05 am

    14. No problem, glad it helped!

      Kalid — August 2, 2007 @ 2:39 am

    15. 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

      sam eismont — August 23, 2007 @ 10:44 am

    16. 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?

      Tom — August 29, 2007 @ 1:49 am

    17. Thanks alot!!!!
      am studying n i have an exam 2mmorow !!
      thnx 4 helpn me

      Besho — September 24, 2007 @ 3:36 pm

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

      Kalid — September 27, 2007 @ 9:23 am

    19. 10 pairs of shoes are well mixed up.4 shoes are randomly picked. What is the probablity of getting at least 1 complete pair

      Sahil — September 30, 2007 @ 12:12 am

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

      Pras — October 2, 2007 @ 6:03 am

    21. 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 :)

      Kalid — October 2, 2007 @ 7:55 am

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

      safa — October 6, 2007 @ 3:32 am

    23. 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..

      safa — October 6, 2007 @ 3:37 am

    24. 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

      Kalid — October 6, 2007 @ 7:58 am

    25. thanks too much
      i understood it
      thanks again

      Anonymous — October 24, 2007 @ 11:38 am

    26. Nice explanation. I was wandering if you could explan some more about COUNTING….

      Thanks..
      splitline..

      splitline — October 30, 2007 @ 8:00 pm

    27. Hi 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.

      Kalid — October 30, 2007 @ 11:06 pm

    28. It was a very useful site,indeed!

      Geetha — November 14, 2007 @ 10:25 am

    29. Thanks Geetha!

      Kalid — November 14, 2007 @ 1:02 pm

    30. Hi
      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

      Hany — November 15, 2007 @ 7:30 am

    31. Hi 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.

      Kalid — November 15, 2007 @ 11:03 am

    32. Thank 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

      Hany — November 16, 2007 @ 10:20 am

    33. didn’t help, im more confused

      Anonymous — November 17, 2007 @ 9:58 pm

    34. Sorry 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.

      Kalid — November 18, 2007 @ 1:26 am

    35. Please help! How many combinations of wins are there in 12 football games?

      Kathleen — November 18, 2007 @ 3:38 pm

    36. Is 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.

      Jon — November 19, 2007 @ 2:59 pm

    37. @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 :)

      Kalid — November 19, 2007 @ 3:09 pm

    38. Thank 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!

      Jon — November 21, 2007 @ 11:57 pm

    39. Hi 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 :)

      Kalid — November 22, 2007 @ 7:59 am

    40. Great 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 :)

      Chris — November 24, 2007 @ 11:31 pm

    41. Hi 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 :) .

      Kalid — November 26, 2007 @ 4:26 pm

    42. I 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?

      murali — November 27, 2007 @ 2:14 pm

    43. Thankz kalid, The site is so very cool.I am glad to visit this.

      Giridhar — November 29, 2007 @ 2:06 pm

    44. Thanks Giridhar, glad you found it :)

      Kalid — November 29, 2007 @ 2:27 pm

    45. 1.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)

      Anonymous — November 29, 2007 @ 6:04 pm

    46. my question is the one above! i need help.. asap thanks

      katie — November 29, 2007 @ 6:19 pm

    47. Thanx a bunch, loved it!

      Naushad Ali — December 1, 2007 @ 5:52 am

    48. @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 :)

      Kalid — December 1, 2007 @ 10:01 am

    49. I 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?

      Dennis — December 1, 2007 @ 1:31 pm

    50. Sorry, THANKS!

      Dennis — December 1, 2007 @ 1:32 pm

    51. Hi 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.

      Kalid — December 2, 2007 @ 1:28 am

    52. Thanks 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!

      Dennis — December 2, 2007 @ 4:10 pm

    53. P.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

      Dennis — December 2, 2007 @ 4:20 pm

    54. Wait 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.

      Dennis — December 2, 2007 @ 5:59 pm

    55. Hi 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 :) .

      Kalid — December 4, 2007 @ 2:15 am

    56. I 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!

      Shakara — December 4, 2007 @ 6:04 pm

    57. Hi 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.

      Kalid — December 5, 2007 @ 4:50 pm

    58. I 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.

      Frank L — December 9, 2007 @ 1:39 am

    59. Sir,

      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

      Benjamin — December 9, 2007 @ 8:36 pm

    60. This 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 Ali — December 13, 2007 @ 4:27 am

    61. I 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?

      Mohammad Ali — December 13, 2007 @ 4:28 am

    62. @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 :) .

      Kalid — December 13, 2007 @ 7:04 pm

    63. Here’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?

      Aztral — December 16, 2007 @ 12:59 pm

    64. Hi 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

      Dennis — December 16, 2007 @ 10:04 pm

    65. Recursive definitions and algorithms. Any suggestions on some links.
      Thanks

      Dennis — December 16, 2007 @ 10:09 pm

    66. @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).

      Kalid — December 16, 2007 @ 10:31 pm

    67. Hi 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 :)

      Aztral — December 18, 2007 @ 7:59 am

    68. Great, 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.

      Kalid — December 18, 2007 @ 11:57 am

    69. I 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

      Jonathan — December 23, 2007 @ 6:21 pm

    70. Hi Jonathan, that’s a bit of a trick question — try doing an example where you pull out 5 random socks and see what happens :) .

      Kalid — December 24, 2007 @ 5:31 pm

    71. Great 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.

      Neil — December 30, 2007 @ 1:40 pm

    72. Can anyone give me the answer to this question.

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

      Tlna — January 1, 2008 @ 10:24 pm

    73. i like it a lot

      Anonymous — January 3, 2008 @ 3:43 pm

    74. i do to

      bob — January 3, 2008 @ 3:43 pm

    75. Here 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.

      Dave — January 4, 2008 @ 9:49 am

    76. Number 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.

      Dave — January 4, 2008 @ 10:43 am

    77. “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.

      DiegoAndresJAY — January 4, 2008 @ 1:31 pm

    78. Hi Diego, thanks for the comment — that’s a nice way to visualize it.

      Kalid — January 5, 2008 @ 12:30 am

    79. Kalid,

      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!

      Matt — January 11, 2008 @ 8:14 pm

    80. Hi Matt, that’s great! The world needs more teachers who focus on more than memorization :) . Good luck!

      Kalid — January 11, 2008 @ 8:25 pm

    81. The 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.

      Topi — January 16, 2008 @ 9:15 pm

    82. I found out the answer(48) but I still cant figure how it was worked out

      Topi — January 16, 2008 @ 9:26 pm

    83. Actually 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.

      Topi — January 16, 2008 @ 10:00 pm

    84. How many 4-letter combinations are there of the letters in c) OUTLOOK?

      Would you please solve this one? Thx.

      Wyner — January 17, 2008 @ 4:15 am

    85. People could skip classes and learn here instead. It’s the same thing, it not better. Great job!

      Michał Lewtak — January 26, 2008 @ 3:01 pm

    86. if*

      Michał Lewtak — January 26, 2008 @ 3:02 pm

    87. Thanks Michal, glad you liked it!

      Kalid — January 26, 2008 @ 4:46 pm

    88. Great stuff, find in great help!

      Rajesh — January 30, 2008 @ 3:47 am

    89. Thanks Rajesh, happy you found it useful.

      Kalid — January 30, 2008 @ 7:52 pm

    90. its a very helpful site for maths students

      naved — January 31, 2008 @ 12:04 am

    91. Thanks naved!

      Kalid — January 31, 2008 @ 12:07 am

    92. The 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

      javad majd — February 27, 2008 @ 2:17 am

    93. thanks 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

      christine — March 3, 2008 @ 7:26 pm

    94. Hang 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?

      Christine — March 3, 2008 @ 7:32 pm

    95. @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).

      Kalid — March 3, 2008 @ 7:47 pm

    96. Hi,
      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.

      Barry G — March 5, 2008 @ 7:21 am

    97. Hi 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).

      Kalid — March 5, 2008 @ 10:20 am

    98. Kalid, thanks a million for your answer - it’s exactly what I was looking for! Keep up the good work!

      Barry G — March 6, 2008 @ 11:19 am

    99. Hi Barry, you’re more than welcome — happy you found it useful.

      Kalid — March 6, 2008 @ 11:33 am

    100. Can 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)

      migz — March 8, 2008 @ 7:21 pm

    101. thanks, I had strep and missed school and this lesson so the site really helped! =^..^=

      Kevin — March 11, 2008 @ 11:48 am

    102. @migz: Sorry, looks like a homework question — try asking your teacher!

      @Kevin: Thanks, glad it was helpful!

      Kalid — March 11, 2008 @ 12:05 pm

    103. really, this is the way one should explain the concepts. thnx a lot, lot, lot……..

      venkatesha prasad — March 22, 2008 @ 2:00 am

    104. You’re welcome Venkatesha, happy you found it useful.

      Kalid — March 22, 2008 @ 7:09 am

    105. i 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

      someone could help me please!! — March 24, 2008 @ 11:03 am

    106. I 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.

      Tina — March 31, 2008 @ 2:00 am

    107. Thanks but I still don’t get it…

      Dominic — March 31, 2008 @ 8:11 pm

    108. @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.

      Kalid — April 1, 2008 @ 12:18 pm

    109. You 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.

      Harley — April 9, 2008 @ 6:39 am

    110. thanks 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!!!

      Neemitha — April 15, 2008 @ 6:08 pm

    111. Awesome Neemitha, I’m glad it worked for you :) .

      Kalid — April 15, 2008 @ 6:37 pm

    112. Heya,
      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?

      Dylan — April 25, 2008 @ 9:12 pm

    113. Hi 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 :) .

      Kalid — April 26, 2008 @ 12:48 am

    114. Help! How many ways can you get at least 7 out of 10 answers correct on a test?

      Carolynn — April 26, 2008 @ 11:56 am

    115. Hi 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)

      Kalid — April 27, 2008 @ 2:27 am

    116. Thanks 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?

      Dylan — April 27, 2008 @ 5:13 am

    117. Hi 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).

      Kalid — April 27, 2008 @ 9:54 am

    118. it 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!!!

      chole — April 29, 2008 @ 2:21 pm

    119. Hi 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 :) .

      Kalid — April 29, 2008 @ 2:28 pm

    120. Hi 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

      Enric — May 2, 2008 @ 1:00 pm

    121. YESSS!! thank god there is a website like this. this is exactly EXACTLY what i wanted.

      THANK YOU!!!

      Benjamin — May 14, 2008 @ 5:52 pm

    122. Thanks Benjamin, I’m glad it helped you!

      Kalid — May 14, 2008 @ 7:47 pm

    123. This is so awsome. I have EOG’s tommorow and I really needed a remider on this stuff, thanks!

      Carly — May 19, 2008 @ 4:58 pm

    124. Excellent, glad it was helpful!

      Kalid — May 19, 2008 @ 10:55 pm

    125. Something 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?

      Dave R. — May 20, 2008 @ 4:52 am

    126. Hi 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.

      Kalid — May 21, 2008 @ 7:51 am

    127. Thank you so much for explaining it in such simple terms. Sure to book mark this link.

      KB — May 27, 2008 @ 10:38 am

    128. You’re welcome KB, glad it was helpful!

      Kalid — May 27, 2008 @ 11:55 am

    129. thnx a lot.i understood the theory clearly only now.again thnx a lot.

      Hari — June 2, 2008 @ 9:52 am

    130. Thanks Hari, glad you enjoyed it!

      Kalid — June 2, 2008 @ 11:05 pm

    131. ur 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

      gerbs — June 4, 2008 @ 3:42 pm

    132. Thanks gerbs, I’m happy it was useful — good luck on your test!

      Kalid — June 5, 2008 @ 11:03 am

    133. Thanks these are very usefull for me

      Sbs Matematik — June 6, 2008 @ 3:58 pm

    134. hi, 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.

      gurneet — June 7, 2008 @ 1:55 am

    135. thanks 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

      taking the SATII — June 7, 2008 @ 6:11 am

    136. It helped. Simple explanation, but effective. The word redundancy in Combination made the things simpler.

      Mahesh — June 8, 2008 @ 11:24 pm

    137. wow!!! 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

      june ace — June 10, 2008 @ 3:21 am

    138. @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!

      Kalid — June 10, 2008 @ 11:07 am

    139. hello, 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

      charbel k. — June 12, 2008 @ 11:39 pm

    140. I ‘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?

      Anonymous — June 18, 2008 @ 6:50 am

    141. I ‘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!)

      Aaronic — June 18, 2008 @ 6:54 am

    142. Thank you very much Kalid

      ankastre — June 19, 2008 @ 1:06 am

    143. Look at permutation this way.. It is about things Per Mutation i.e. all possible ways things can take shape. :) What say?

      Sangramsinh Takmoge — June 22, 2008 @ 4:35 am

    144. UMMM….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

      JUNE — June 26, 2008 @ 8:34 am

    145. Thank you so much kalid…

      program images webmaster — June 30, 2008 @ 10:31 am

    146. yep its rght the boy has keot 1 pesos

      Anonymous — July 6, 2008 @ 11:47 pm

    147. This is one fantastic website
      kudos to u guys
      keep up the gr8 work

      Sid — July 9, 2008 @ 12:29 pm

    148. 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….

      seviye belirleme sınavı — July 16, 2008 @ 5:33 am

    149. Wow! Excellent explanation. I really understood this.

      anonymous — July 16, 2008 @ 8:29 am

    150. Permutations and combinations - never thought I would understand them.

      blackstar104 — July 16, 2008 @ 8:30 am

    151. Thanks alot for sharing it across. Soperb explanation!

      Regards,
      Sharma

      Sharma — July 27, 2008 @ 7:16 am

    152. Thanks alot for sharing! Superb explantion.

      Regards,
      Sharma

      Sharma — July 27, 2008 @ 7:17 am

    153. 1. 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

      Sharma — July 27, 2008 @ 11:12 am

    154. How many 6 digit combimations are ther from 17 through 47? Can you send me them? Thanks

      Rick — July 28, 2008 @ 6:26 pm

    155. I 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?

      Prat Shell — July 30, 2008 @ 5:55 pm

    156. Hi 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.

      Kalid — July 30, 2008 @ 6:11 pm

    157. @Sharma: Whoops, thanks for the correction! I’m fixing it now.

      Kalid — July 30, 2008 @ 6:12 pm

    158. Many 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? :-)

      Prat Shell — July 30, 2008 @ 6:19 pm

    159. @Pret: Glad you enjoyed it! I’ve started thinking about making quick videos, maybe I’ll try that out :) .

      Kalid — July 31, 2008 @ 3:45 pm

    160. hi–
      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

      antonio — August 9, 2008 @ 10:01 am

    161. Hi,
      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,

      Antonio — August 9, 2008 @ 11:20 am

    162. how 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”

      Ashu — August 19, 2008 @ 12:32 pm

    163. Your 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.

      Pradeep — August 22, 2008 @ 9:32 am

    164. Great 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?

      Mark — August 27, 2008 @ 9:40 pm

    165. Ok, 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.

      CS — September 5, 2008 @ 10:20 am

    166. I 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 Marks — September 13, 2008 @ 1:50 pm

    167. Never 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. 10×10x10 gives 1000 possible lock settings using 3 dials having 10 numbers on each dial. Thanks.

      Sari Marks — September 13, 2008 @ 2:03 pm

    168. i 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!!

      priya — September 14, 2008 @ 5:51 am

    169. @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!

      Kalid — September 16, 2008 @ 6:20 pm

    170. u really did help me more than my maths teacher! rawk on dude…

      ljw — September 28, 2008 @ 2:54 am

    171. @ljw: Great, glad you enjoyed it!

      Kalid — September 28, 2008 @ 11:41 am

    172. if 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?

      Matt — September 29, 2008 @ 8:49 pm

    173. Thanx alot Kalid :)
      you actually explain it better than my teacher

      Mahmoud Hesham — October 8, 2008 @ 11:52 pm

    174. @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!

      Kalid — October 9, 2008 @ 4:49 pm

    175. Great, Thx…

      transpalet — October 13, 2008 @ 6:32 am

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

      online haber — October 16, 2008 @ 6:28 am

    177. I 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”.

      anisa — October 20, 2008 @ 7:56 am

    178. Hi Anisa, a better place for this type of specific question is probably http://mathforum.org/dr.math/.

      Thanks,

      -Kalid

      Kalid — October 20, 2008 @ 11:46 am

    179. Thank you very much Kalid!!

      have a nice day,

      anisa — October 20, 2008 @ 2:09 pm

    180. Hi 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

      Alex Cameron — October 21, 2008 @ 11:58 am

    181. Finally 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.

      Ayush — October 23, 2008 @ 8:30 am

    182. @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.

      AB — October 30, 2008 @ 4:17 pm

    183. Alex, 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).

      Anonymous — November 1, 2008 @ 10:46 am

    184. thanx,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”…

      crystal — November 11, 2008 @ 8:59 am

    185. IT IS AN AMAZING MATHS . I WISH TO ANSEW THE QUESTON.BECAUSE IT GIVES ME ENCOURAGEMENT OF DOING IT THANKS.

      MERCY — November 12, 2008 @ 2:41 pm

    186. My 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?

      Derek — November 16, 2008 @ 12:50 am

    187. My 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

      Laura — November 16, 2008 @ 2:05 am

    188. @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).

      Kalid — November 16, 2008 @ 6:06 pm

    189. @ 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.. :(

      Abdulla — November 23, 2008 @ 12:38 pm

    190. Hi 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”.

      Kalid — November 23, 2008 @ 2:42 pm

    191. distribute 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

      kalyan — November 25, 2008 @ 9:04 am

    192. permutations 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.

      sadie — November 25, 2008 @ 1:41 pm

    193. On 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?

      sassa — December 7, 2008 @ 11:21 pm

    194. mind blowing

      santosh — December 8, 2008 @ 5:22 am

    195. thanks man, helps a bit

      Deep — December 10, 2008 @ 9:14 pm

    196. hi 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

      Anonymous — December 14, 2008 @ 3:29 pm

    197. hi 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

      tabi — December 14, 2008 @ 3:29 pm

    198. I 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?)

      Aaron Willhite — January 6, 2009 @ 6:13 pm

    199. Hi 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?

      Kalid — January 6, 2009 @ 7:53 pm

    200. still don’t know about it…
      I don’t understand clearly..

      Snow Zhao — January 8, 2009 @ 4:22 pm

    201. Wonderful 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

      melvin — January 9, 2009 @ 3:28 am

    202. @ 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.

      ctlr — January 9, 2009 @ 2:44 pm

    203. Wow, this helped me a lot! thanks!

      Ann — January 19, 2009 @ 6:12 am

    204. Problem:- 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

      Vibhanshu Bhardwaj — January 19, 2009 @ 10:29 pm

    205. at last i knew it, this was clearer than that of what our prof explained..
      wHeew!
      thanks a Lot!

      sandy — February 1, 2009 @ 7:30 am

    206. Awesome 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

      Nouri — February 3, 2009 @ 8:28 pm

    207. @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 :) .

      Kalid — February 3, 2009 @ 9:25 pm

    208. amazing how you explain the question

      Anonymous — February 6, 2009 @ 10:45 am

    209. im 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……….

      Anonymous — February 6, 2009 @ 10:51 am

    210. Thanks..great forum.

      How many 4-letter combinations are there in the word, OUTLOOK?..thnx — February 9, 2009 @ 7:43 am

    211. My 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

      Sandy — February 10, 2009 @ 10:04 am

    212. when 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

      Mathtut — February 12, 2009 @ 9:03 pm

    213. Thanks so much!!

      Sandy — February 13, 2009 @ 3:32 pm

    214. @Mathtut: Thanks for helping out Sandy!

      Kalid — February 13, 2009 @ 4:23 pm

    215. stupid

      cory — February 15, 2009 @ 7:13 pm

    216. when you’re doing something, permutations, or combinations, and order DOES NOT matter, it becomes a combination, right?

      CourtneyAllison — February 18, 2009 @ 2:59 pm

    217. comment 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.

      CourtneyAllison — February 18, 2009 @ 3:02 pm

    218. can 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?

      chizen — February 19, 2009 @ 10:09 am

    219. thanks a lot, this is way simpler than the explanations from my book *chuckles*

      Regards,
      Cady

      cady — February 24, 2009 @ 4:53 pm

    220. *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

      cady — February 24, 2009 @ 5:20 pm

    221. Help 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.

      Ekum — March 2, 2009 @ 6:18 am

    222. My god you are a genius.. do you know how many time I read my probability book to understand permutation and combination concept.

      Biniyam — March 2, 2009 @ 5:05 pm

    223. @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!

      Kalid — March 2, 2009 @ 7:35 pm

    224. A 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

      Layla — March 4, 2009 @ 4:44 pm

    225. oh 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

      Layla — March 4, 2009 @ 4:54 pm

    226. @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.

      Chetan — March 6, 2009 @ 4:19 pm

    227. I’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.

      Chetan — March 6, 2009 @ 4:36 pm

    228. Well 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.

      Anonymous — March 8, 2009 @ 2:00 pm

    229. Thanks ‘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.

      Chetan — March 8, 2009 @ 6:44 pm

    230. Yes 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.

      Sreshtha — March 11, 2009 @ 4:10 am

    231. I 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?

      Cathleen — March 11, 2009 @ 5:17 pm

    232. You are providing an exceptionally well done service to anyone trying to really understand mathematics, Thank You!

      CT — March 12, 2009 @ 4:29 am

    233. This a great website! It explains everything so well. I rathe do this than go to school!

      Unshu — March 12, 2009 @ 4:59 pm

    234. Excellent.

      hiteshgarg7 — March 13, 2009 @ 4:53 am

    235. I don’t get why it is 8*7*6

      J — March 13, 2009 @ 6:20 pm

    236. I don’t get why they are multiplied at all

      J — March 13, 2009 @ 6:59 pm

    237. @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.

      Kalid — March 13, 2009 @ 7:33 pm

    238. Man 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

      j — March 13, 2009 @ 8:21 pm

    239. whoaa no, there I have it. You can actually picture it as a cube. I was lost there for awhile. I really appreciate your help!

      j — March 13, 2009 @ 8:27 pm

    240. sorry i meant: cube isn’t the proper word because it isn’t square.

      thanks again

      j — March 13, 2009 @ 8:31 pm

    241. THANK 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… :)

      Miguel Angel — March 21, 2009 @ 7:33 pm

    242. @Miguel: You’re welcome! Really glad it helped :)

      Kalid — March 22, 2009 @ 12:04 am

    243. man this site is freakin awsome

      i — March 24, 2009 @ 7:25 pm

    244. @i: Glad you liked it!

      Kalid — March 24, 2009 @ 10:36 pm

    245. Hello, 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

      Jordan — March 27, 2009 @ 3:06 pm

    246. @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.

      Kalid — March 27, 2009 @ 3:21 pm

    247. Thanks 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.

      Jordan — March 27, 2009 @ 4:15 pm

    248. Is that what the answer is supposed to be 15? I would have thought 210. I am really lost on this one.

      Jordan — March 27, 2009 @ 9:36 pm

    249. Hi 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 :) .

      Kalid — March 28, 2009 @ 10:36 am

    250. Also, 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.

      Kalid — March 28, 2009 @ 10:41 am

    251. Hello 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

      Jordan — March 28, 2009 @ 2:32 pm

    252. sorry that should have been x= right y= up

      Jordan — March 28, 2009 @ 2:49 pm

    253. You opened my eyes. Finally i can see the difference between Permutations and combinations.

      God Bless.

      Anonymous — March 29, 2009 @ 10:00 am

    254. @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!

      Kalid — March 29, 2009 @ 2:43 pm

    255. How 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

      Jim — March 30, 2009 @ 8:43 pm

    256. hello, 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.

      lind — April 2, 2009 @ 10:22 am

    257. I 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.

      Lola — April 2, 2009 @ 5:55 pm

    258. If 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

      Jordan — April 2, 2009 @ 5:59 pm

    259. I 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?

      Kurt — April 17, 2009 @ 10:17 am

    260. you 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!

      Anonymous — April 26, 2009 @ 4:42 pm

    261. why 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!!

      Jorge (GEORGE) — April 26, 2009 @ 4:57 pm

    262. thanks!! i have a test tomorow and this made sense! ‘’)

      reed — May 5, 2009 @ 6:34 pm

    263. @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! :)

      Kalid — May 5, 2009 @ 6:49 pm

    264. hey 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 :)

      Anonymous — May 7, 2009 @ 3:31 pm

    265. Thanks, that helped a lot . Much simpler than Wikipedia.

      Steven — May 9, 2009 @ 8:59 pm

    266. this is cool som what understood helping me for the end of grade test!! :=) :-}

      why but ayways thanks — May 12, 2009 @ 4:14 pm

    267. Im 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

      laura — May 14, 2009 @ 12:46 pm

    268. Would you want to go into greater detail ?
      Especially combinations ? There are so many more formulas than the ones stated.
      But thanks anyway.

      Soham — May 18, 2009 @ 8:35 pm

    269. @Soham: You’re welcome — I’m planning on doing more follow-up articles on more advanced types of counting. Thanks dropping in.

      Kalid — May 18, 2009 @ 10:29 pm

    270. hua!! it’s so confusing.. >.

      claudia lidwina — May 21, 2009 @ 4:45 am

    271. It’s really very helpful.
      this article justifies the betterexplained.com
      best.
      Other advanced topics are expected.

      Thanx!

      gurudeep — May 26, 2009 @ 8:33 am

    272. This was really helped clarify all my concerns about combinations and permutations. thank you so much. my teacher did a very shoddy job.

      Anonymous — May 28, 2009 @ 10:23 am

    273. @Gurudeep, Anonymous: Thanks!

      Kalid — May 28, 2009 @ 12:28 pm

    274. Hi, 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?

      Ashley Dudzinski — May 31, 2009 @ 9:33 pm

    275. @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!

      Kalid — June 1, 2009 @ 12:15 am

    276. Thanks, I have a test tomorrow and this really cleared it up

      Katie — June 1, 2009 @ 6:11 pm

    277. @Katie: Great, good luck!

      Kalid — June 1, 2009 @ 6:40 pm

    278. Greetings,

      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 chanhankiang21@hotmail.com

      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 (9×9x8×7)+(26×25x24). 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 4×5x5×5x5 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

      chanhankiang — June 6, 2009 @ 2:10 am

    279. whoah… 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 =)

      Christy — June 18, 2009 @ 7:34 pm

    280. oh, 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…

      Christy — June 18, 2009 @ 7:41 pm

    281. @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!

      Kalid — June 18, 2009 @ 10:28 pm

    282. Thanks admin

      2009 Sbs Sonuçlari — June 21, 2009 @ 3:19 am

    283. this 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.

      henry gulani — June 24, 2009 @ 10:17 am

    284. khjhjhkj

      Anonymous — June 26, 2009 @ 10:22 am

    285. Thanks admin..

      Sbs sonuçları 2009

      Günlüğüm

      Sbs Sonuçları — June 29, 2009 @ 12:49 pm

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