GMAT Permutations and Combinations


A permutation is a possible order in which to put a set of objects.  Suppose I had a shelf of 5 different books, and I wanted to know: in how many different orders can I put these 5 books?  Another way to say that is: 5 books have how many different permutations?

In order to answer this question, we need an odd math symbol: the factorial.  It’s written as an exclamation sign, and it means: the product of that number and all the positive integers below it, down to 1.  For example, 4! (read “four factorial”) is

4! = (4)(3)(2)(1) = 24


Here’s the permutation formula:

# of permutations of n objects = n!

So, five books right the number of permutations is 5! = (5)(4)(3)(2)(1) = 120



A combination is a selection from a larger set.  Suppose there is a class of 20, and we are going to pick a team of three people at random, and we want to know: how many different possible three-person teams could we pick?  Another way to say that is: how many different combinations of 3 can be taken from a set of 20?

This formula is scary looking, but really not bad at all.  If n is the size of the larger collection, and r is the number of elements that will be selected, then the number of combinations is given by

# of combinations = {n!}/{r!(n-r)!}


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Again, this looks complicated, but it gets simple very fast.  In the question just posed, n = 20, r = 3, and n – r = 17.  Therefore,

# of combinations = {20!}/{3!(17)!}


To simplify this, consider that:

20! = (20)(19)(18)(17)(the product of all the numbers less than 17)


Or, in other words,

20! = (20)(19)(18)(17!)


That neat little trick allow us to enormously simplify the combinations formula:

# of combinations =  {(20)(19)(18)(17!)}/{3!(17)!}={(20)(19)(18)}/{3!}={(20)(19)(18)}/{(3)(2)(1)}=1140

That example is most likely harder than anything you’ll see on the GMAT math, but you may be asked to find combinations with smaller numbers.

Practice Questions

1) A bookseller has two display windows.  She plans to display 4 new fiction books in the left window, and 3 new non-fiction books in the right window.  Assuming she can put the four fiction books in any order, and separately, the three non-fiction books in any order, how many total configurations will there be for the two display windows?

    (A) 24
    (B) 72
    (C) 144
    (D) 336
    (E) 420


2) The county-mandated guidelines at a certain community college specify that for the introductory English class, the professor may choose one of three specified novels, and choose two from a list of 5 specified plays.  Thus, the reading list for this introductory class is guaranteed to have one novel and two plays.  How many different reading lists could a professor create within these parameters?

    (A) 15
    (B) 30
    (C) 90
    (D) 150
    (E) 360

Answers and Explanations

1) The left window will have permutations of the 4 fiction books, so the number of possibilities for that window is

permutations = 4! = (4)(3)(2)(1) = 24


The right window will have permutations of the 3 non-fiction books, so the number of possibilities for that window is

permutations = 3! = (3)(2)(1) = 6


Any of the 24 displays of the left window could be combined with any of the 6 displays of the right window, so the total number of configurations is 24*6 = 144

Answer: C.

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2) There are three possibilities for the novel.   With the plays, we are taken a combination of 2 from a set of 5 right  n = 5, r = 2, n – r = 3

# of combinations = {5!}/{2!3!} = {(5)(4)(3)(2)(1)}/{(2)(1)(3)(2)(1)} = {(5)(4)}/2 = 10


If the plays are P, Q, R, S, and T, then the 10 sets of two are PQ, PR, PS, PT, QR, QS, QT, RS, RT, & ST.

Any of the three novels can be grouped with any of the 10 possible pairs of plays, for a total of 30 possible reading lists.

Answer: B.

Special Note:

To find out where permutations and combinations sit in the “big picture” of GMAT Quant, and what other Quant concepts you should study, check out our post entitled:

What Kind of Math is on the GMAT? Breakdown of Quant Concepts by Frequency

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28 Responses to GMAT Permutations and Combinations

  1. NAR June 25, 2016 at 11:40 pm #

    Hi Mike 🙂
    Why the use of combination or permutation is needed to perform the business function?

    • Magoosh Test Prep Expert
      Magoosh Test Prep Expert June 27, 2016 at 6:03 am #

      Hi NAR,

      Honestly, I believe that these types of questions are intended to thoroughly test the problem solving skills of test takers rather than measure a real skill that will be needed in business! 🙂

  2. MICHAEL December 10, 2015 at 2:02 am #

    Hi Mike,

    I still don’t understand why #1 we use permutation when it say clearly that ” in any order” while #2 we use combination when the question stated in a way I believe order of arrangement matter. I have read your post regarding to permutation and combination already. I am just very confused. thanks for your help

    • John August 26, 2016 at 9:23 am #

      Hai…I would say that both the problems use permutation. I the case of #1, order and repetition is not a concern. i.e. from the list of ABCD, AB and BA are same combinations. U eat a bread and a tea or a tea and bread are same.

      While in #2, from the list of 5 plays, we need only 2 plays. U can take 1,2 or 2,3 or 3,4 or 4,5 or 1,3 or 1,4 or 1,5, or 2,4 or 2,5 or 3,5. These are the various distinct combinations.
      All the above combinations have mirror pairs such as 2,1 or 3,2 or 4,3 or 5,4 etc which are mere repetitions.We weed out these repetitions using the nCr formula.

    • David October 5, 2016 at 8:16 am #

      I think the key word to note in Question #1 is “configurations”. This really means permutations. The phrase “in any order” just describes how she is able to arrange the books, with no bearing on how we decide to count the permutations or combinations.

      As for Question #2, I agree that the wording is a bit ambiguous.

  3. Neeraj September 20, 2015 at 10:27 am #

    Thank you Magoosh.
    The second question can also be solved in a simple way

    # of combinations of novels + # of combinations of plays
    = 3!/1*2! + 5!/2!*3!
    = 3 * 10
    = 30

    • Alex September 29, 2015 at 8:52 pm #


      Equation should be

      = 3!/1*2! * 5!/2!*3!

      It is multiplication of the two combinations not the addition of the two.

  4. siddhartha September 16, 2015 at 7:36 pm #

    Very helpful posts. However, I have a doubt. Although I kind of understand why in the second question, you went with 3C1 * 5C2 = (3*10)= 30, I am confused about one thing.

    Another way to do solve this problem would be to make three boxes– first box for the novel, second for a play, third also for a play.

    The first box can be filled in 3 ways; the second in 5 ways, the third in 4 ways.

    So total no. of combinations: 3*5*4 = 3*20=60. It gives us a different answer. Why is this wrong? Am I missing something about “order matters, order does not matter”.

    I suppose in this case, order does not matter. But I don’t know how to translate that into the Math here. Furthermore, I consulted your earlier post on P&C ( In the Shakespeare example, where you say:

    “For example: Shakespeare wrote fifteen comedies, ten histories, and twelve tragedies. If we are going to pick one of each kind, and ask how many different trios of plays can we create, the total number is simply 15*10*12. ”

    How is the Shakespeare problem above different from this question here? I would be most grateful for your explanation.

    • Mike MᶜGarry
      Mike McGarry September 17, 2015 at 3:11 pm #

      Dear Siddharta,
      I’m happy to respond. 🙂 This is a very tricky think about counting problems. Your method was perfectly valid, but you overlooked one thing: repetitions.
      Let’s call those five Shakespeare plays {A, B, C, D, E}. You said: five choices for the first play, and four choices for the second play. That’s perfectly true, but each pair gets counted twice. If we pick A first, and B second, or if we pick B first, and A second, either results in the same pair (A, B) on the reading list. Order doesn’t matter here, but in your calculation, you presumed that order mattered, so you wound up double-counting, which gave you an answer twice as big as the correct answer. All of these issues are explained in detail in the Magoosh lessons on counting.
      Does all this make sense?
      Mike 🙂

  5. Ally August 18, 2015 at 9:04 am #

    Hi Mike,

    I’ve come across your blog 24 hours before my gmat. and it feels godsent. Just reading through these simple explanations with bits of humor thrown in is calming me a lot 🙂
    I had one simple question- I get confused about differentiating between permutation and combination sometimes. Is there a blog post explaining that?
    THanks so much

  6. Saakshi August 10, 2015 at 10:01 am #

    I’m so grateful to this site. Not that i have secured a score yet, but the prep time is so much fun. thanks to magoosh.
    such little details explained so explicitly. sometimes I couldnt even fathom where i was going wrong. thanks to you, i realized depending too much on formulas too can be an error.
    thankyou again! keep up!

  7. kimberly October 8, 2014 at 2:55 am #

    what are example of companies that uses permutation or combination to perform their inventory management or other business function

    • Mike MᶜGarry
      Mike October 8, 2014 at 10:29 am #

      Dear Kimberly,
      I’m happy to respond. 🙂 First of all, understand that you are asking the wrong person. By trade, I am a GMAT expert: I can tell you cartloads about how this math appears on the GMAT. I am NOT an industry specialist in any industry, so I am certainly not equipped to discuss details of how this math might be used in day-to-day operations of specific businesses. I have my guesses, but I have no data at my fingertips on that.
      Here’s what I will say. I could easily imagine the following scenario playing out many times. The CEO or CFO or other high muck-a-muck at a corporation wants to know how many ways they can market something or package something, or some similar question — what we would recognize as a combinations question. The high muck-a-muck doesn’t know this, so that person sends the question to some underling, who asks someone else, and eventually the question finds it way to someone such as an engineer who realizes that it’s actually a very easy question to answer when one knows the math. This engineer produces the correct answer, and this answer migrates back up the food chain until it reaches the high muck-a-muck who asked in the first place. Now, here’s the funny thing. If we surveyed the company, that is, if we asked the high muck-a-muck or one of the immediate underlings, “Do you use permutations and combinations?”, their answer might well be: “What are those? I’ve never even heard of them!” Even though someone at the company actually did use permutations and combination to answer the question, the folks who received the answer may not be aware of that in the least. People who run companies and their press agents often have lots of numbers at their disposal, and they don’t always understand exactly how every last number was calculated: after all, knowing that calculation is not part and parcel of their job. All this would make it extremely problematic to determine which companies are actually using this math.
      All this is to say that is you have all the math at your fingertips, you will walk through the corporate works with a kind of magic power, able to compute things quickly that baffle others.
      Does all this make sense?
      Mike 🙂

  8. Herpal Pabla August 23, 2014 at 12:41 pm #

    Greetings Mike,
    I wanted to thank you for the great work and explanations you have provided us. I have been studying for the GMAT for a while, and there are some concepts in my prep book that don’t thoroughly explain on how to get certain types of questions. The addition of your material has def helped in a significant way.
    Kind Regards,
    Herpal Pabla

    • Mike MᶜGarry
      Mike August 23, 2014 at 12:49 pm #

      Dear Herpal,
      My friend, you are more than welcome. 🙂 I am very glad to hear that you found this helpful! Best of luck to you in the future!
      Mike 🙂

  9. shilpi February 20, 2014 at 11:17 am #

    Could you please provide some more practice questions based on the above concept. Thanks!

  10. Steven July 30, 2013 at 8:27 am #

    “With the plays, we are taken a combination of 3 from a set of 5”

    Should be 2 from a set of 5. Otherwise, a very good post.

    • Mike MᶜGarry
      Mike July 30, 2013 at 9:58 am #

      Thanks for catching that — I just fixed it. Incidentally, I probably made the mistake because, as you may know, 5C2 = 5C3 — at a certain level of analysis, there’s no difference picking two from five or three from five. But for clarity, it’s good to make this change. Thanks,
      Mike 🙂

  11. Naren May 6, 2013 at 7:16 pm #

    Hi Mike

    Can we do it this way?

    1/3 * 2/5 * 1/4 = 1/30

    so we have 30 combination?


    • Mike MᶜGarry
      Mike May 7, 2013 at 9:47 am #

      I assume you mean practice question #2. Yes, that would be another way to compute the number of possible reading lists.

  12. Rahul Sehgal March 24, 2013 at 11:00 am #

    Mike – Thanks for the valuable information. I have a clarification to make with respect to question 1.

    I was thinking – would not the answer be 24 + 6 = 30. I am thiking as she wants the ‘ficiton’ books still too be placed on the left hand side in any order and the same goes for ‘non fiction books.

    Am i missing a trick here ?

    Rahul Sehgal
    GMAT aspirant !!

  13. Gopal September 10, 2012 at 1:40 pm #

    A nice intro to P&C and, probably you are true WRT the GMAT, this is probably more than enough to crack

    • Mike MᶜGarry
      Mike September 10, 2012 at 5:21 pm #

      Thank you very much.
      Mike 🙂

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