Three GRE Challenge Combinations and Permutations Problems

These challenge questions are great for extra practice. However, be warned: while I provide the answers, I don’t provide the explanations. If you’d like to try practice problems with explanations, check out Magoosh’s 15 GRE math practice problems and my biweekly Brain Twisters!

Over the last few weeks, I have gone through GRE combinations and permutations. My goal is to make this formidable concept less intimidating. If you’ve watched the videos and feel comfortable with the math, I would say continue practicing but do not worry too much about this problem type. You will probably see one, though you will unlikely see two, on the actual test. So your time will better be spent on other concepts, such as work rate, sets, series, and coordinate geometry. (I will be covering these concepts soon!). Unless, that is, you are aiming for the 160+ category/the top 90% (it’s great that I can finally speak of the Revised GRE in terms of the Revised GRE scoring). The combinations and permutations I covered, while tough, may not be as difficult as a question you may see on the test, if ETS decides to make one of its most difficult questions a combinations and permutations one. Thus I have decided to devise some pretty diabolical questions. In fact, the last two are so convoluted and tricky that they wouldn’t even show up on the test. Again, this is only for those who are already comfortable with combinations and permutations, and who are going for the near perfect score. Otherwise, this could be a waste of your time (of course the less practical side of me thinks that this level of problems is fun!).


Warming Up An equal number of juniors and seniors are trying out for six spots on the university debating team.  If the team must consist of at least four seniors, then how many different possible debating teams can result if five juniors try out?

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(A)  50 (B)  55 (C)  75    (D) 100  (E)  250


Difficulty Level: Hard A bibliophile plans to put a total of seven books on her marble shelf. She can choose these seven books from a mixture of works from Antiquity and works on Post-modernism, of which there are seven each. If the shelf must contain at least four works from Antiquity, and one on Post-Modernism, then how many ways can he select seven books to go on the shelf?

(A)    441 (B)    1225 (C)    1666    (D)   1715  (E)    1820


Difficulty Level: Are you kidding me?!? An artist is planning on mixing together any number of different colors from her palette. A mixture results as long as the artist combines at least two colors. If the number of possible mixtures is less than 500, what is the greatest number of colors the artist could have in her palette? (A)  8 (B)  9 (C)  11  (D) 12  (E)  13


Check out the answers: 1. B   2. D   3. A


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20 Responses to Three GRE Challenge Combinations and Permutations Problems

  1. ram January 31, 2014 at 2:52 am #

    Refering to warming up Q, I didn’t get y u didn’t consider 6seniors and no juniors as another case.

    Conditions mentioned are
    -Equal no of J and S are taking part.. Donno how many and could be any..
    -Atleast 4S in a team.

    And assumption of 6S of 6J and going through process is not violating any of the mentioned rules. So Y not 6S and 6J but only 5S and 5J is the case?

    • Chris Lele
      Chris Lele January 31, 2014 at 1:30 pm #

      Hi Ram,

      Yes, that is real tricky! It seems like I missed something, but if you look at the text it says an equal number of juniors and seniors try out–this number equal 5 (“if five juniors try out”). That means exactly 5 seniors try out. Therefore, there can be no case in which 6 seniors try out, since there are only 5 seniors trying out.

      Hope that clarifies things!

  2. Connie December 16, 2013 at 5:57 am #

    Hi, I don’t understand the last question, could you please explain for me? Thank you so much!!

    • Chris Lele
      Chris Lele December 16, 2013 at 11:51 am #

      Yes, that is a toughie!

      The formula for the total number of ways you can combine n elements from a group of x elements, is 2^x – 1. With this question, there have to be at least two members, so we have to subtract x from the equation above. That last part is not too important because you just want to get a sense of what number for x, when plugged into the equation above, will give us close to 500. 2^9 = 512. So though that makes it seem like (B) would be the answer, remember that you have to have fewer than 500 colors. Even with the 9 that we subtract from 512 (remember we have to subtract each individual color because it doesn’t count as a mixture), we are still over 500. Therefore, 8 is the greatest number of colors that the artist could have in his palette.

      Hope that helps!

  3. David March 3, 2013 at 9:54 am #

    Hi Chris,

    For the 2nd question, doesn’t order matter? Shouldn’t it be a permutation? If we’re arranging books on a bookshelf, it matters if a book is all the way at the end, or the middle, or the beginning, etc.

    What am I missing here?

    • Chris Lele
      Chris Lele March 5, 2013 at 1:55 pm #

      Hi David,

      For this question, the key is noticing the words “select seven books to go on the shelf.” Once the books are on the shelf, the order doesn’t matter. Thus no permutations formula.

      Hope that helps :).

  4. maulik March 2, 2013 at 11:54 pm #

    In the 2nd question, what is wrong if i am doing this way?

    1st select 4 books of Antiquity in 7C4 ways

    then select 1 book of Post-modernism in 7C1 ways

    Finally select 2 books from remaining 9 books (3 of A and 6 of PM) in 9C2 ways

    Then multiplying all of them.

    So, what I am doing wrong here?

    • Chris Lele
      Chris Lele March 5, 2013 at 2:03 pm #

      Hi Maulik,

      I think it’s the last part where you went wrong. When you select the remaining two books you are multiplying that answer by the number of possible groupings of the Antiquity books and the PM books. However, there are some redundant cases, so you will end up counting. To avoid this you want to treat each instance discretely.


      Use the combination formula for each of the three cases and then add the cases together.

      Hope that makes sense!

  5. neha July 9, 2012 at 10:46 am #

    Its really nice to read all your posts. Can ETS ask questions about cyclic permutation?

    • Chris Lele
      Chris July 9, 2012 at 3:30 pm #

      Glad you like the posts :).

      To the best of my knowledge there are no cyclic permutations on the GRE. That falls under higher-level math not tested on the GRE.

      Hope that helps 🙂

  6. Aman April 11, 2012 at 4:23 am #

    Moreover ,the language introduction and question length also adds up to give it a pusillanimous effect ….

    • Chris Lele
      Chris April 11, 2012 at 2:52 pm #

      Yes, I think you mean the set up with the artist’s palette. What do you mean by ‘pusillanimous’? It means timid, lacking courage. Perhaps you meant that the question is so daunting as to make you feel pusillanimous. Hopefully, test day you will feel doughty 🙂 (though not doubt-y).

  7. Aman April 11, 2012 at 4:20 am #

    Hi Chris,
    These are great problems rather i would call them mind activating one,…..I was able to do all of them correctly(Though a silly thing made my 2nd answer wrong)

    In the 2nd question i used exactly the same method as shown in the video but found out my answer to wrong ,then after watching the video i realized ,the problem was the old enemy of mine,,,……….CALCULATION i missed out a 7X5 while noting values from my step.(Sometimes feel like punishing myself for things like these)

    In the third question i used a method somewhat same as yours but tried with value 8 and got my answer.

    The third one can be a very daunting in exam (IF IT ETS WISH TO EXPERIMENT ON US)

    Thanks for uploading them

    • Chris Lele
      Chris April 11, 2012 at 2:50 pm #

      Hi Aman,

      You are definitely welcome. Yes, “if ETS tries to experiment on us…” Number 3 is quite a doozy. Knowing that the sum of all the combinations of x is equal to 2^x – 1 makes the question far more manageable 🙂

  8. Praveen January 5, 2012 at 8:23 am #


    Y is this method i tried wrong?

    Since we HAVE to choose 4-A and 1-PM , i worked like this..

    case 1) 4-A 1-PM 2-A = 7C4 x 7 x 3C2 (3 being the remaining number of A books)

    case 2) 4-A 1-PM 2-P = 7C4 x 7 x 6C2 (choosing 2 from 6 remaining P books)

    case 3) 4-A 1-PM 1-A 1-PM = 7C4 x 7 x 3C1 x 6C1

    but the answer turned out to be very huge.. is it because i chose the mandatory condition as a fixed set?

    • Chris Lele
      Chris January 5, 2012 at 2:54 pm #


      I think the problem you are running into is you are coming up with a new group, ‘P’, whereas that should be part of the PM group.

      • Steffi July 14, 2013 at 5:01 am #

        what Difference does it make in the logic if we consider that way ?

        • Chris Lele
          Chris Lele July 15, 2013 at 2:34 pm #

          See, by using the combinations formula twice for the Antiques, we are treating the Antiques as separate, as though there is an Antique1 and an Antique 2 group, which would dramatically increase the number of possible groups. Therefore, it is important not to break up a single group (i.e., Antiques) into mini-groups.

          Hope that makes sense!

  9. Egor January 4, 2012 at 6:36 am #

    Dear Chris,

    there is a quick way for Q3.
    using a standard formula for the total number of combinations, which is 2^n, you can solve the question like this:

    2^8=256. however, we don’t need the total number and have to exclude 8C0 and 8C1 (these are not mixtures). 256-8-1=247500.

    the correct answer is 8

    • Chris Lele
      Chris January 5, 2012 at 2:52 pm #

      Hi Egor,

      Yes, you are right. We are basically adding up the numbers in Pascals triangle. In this case 2^n is fastest, where n is the number of colors. I filed this problem under combinations/permutations, but as that is not the most efficient way of doing it, it technically doesn’t fit.

      Thanks for your insight. One quick thing, I think you meant 247 not 247500. Also, it is a good idea to test 2^9 as well, which is 512 – 9 = 503.

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