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The Difficulty of Context: Combinations and Permutations Questions

Over the last couple of months, I have covered a fair amount of GRE combinations and permutations questions on the blog. I’ve gone from the very basic to the very difficult (so difficult that they are more ‘fun problem’ than practical practice).

Often, when you are stumped on a problem, it is not that the math is difficult, but that the context may be unfamiliar.

Combinations/permutations is an excellent example of this phenomenon. If you’ve been watching my videos on this topic, you should be very comfortable with the math. But that doesn’t mean that on test day you will be able to answer a combinations/permutations question correctly. To demonstrate, I will give you the following questions:

1. How many groups of three can you form if you choose from a total of seven?

2. How many ways can you sit seven students in three chairs (one chair per student)?

3. How many different triangles can you form by connecting the vertices of a heptagon?

4. A doctor is to administer pyroxeline, damalnethamide, and cyrotinin to his patients, a treatment plan that is to take place over the course of the week. If he can administer each drug once and cannot give more than one drug a day, how many unique schedules of medications can he create, if the order in which he administers the drugs is important?


While these questions may seem to be in wildly divergent contexts, each question can be answered with either 7C3 or 7P3. Indeed, I could have actually included a dozen more examples, all of which have very different contexts. The key – besides lots of practice  – is learning the best way to think about how to solve problems, and not to rely on the problems surface appearance to inform your approach.

For instance, do not think that all polygon questions that talk about joining vertices call for a similar approach to the one above. Likewise, if a question talks about a doctor administering different drugs, don’t think, oh yeah, I remember that on Chris’ blog post, and he used the permutations formula, not the combinations formula. Your goal should be to understand why that specific problem employed permutations, and not combinations.

The test writers are very aware of the vast number of possible contexts. And, while the math won’t change too much, they will often wrap familiar math concepts in misleading guises. Cracking the problem will be more a matter of choosing the correct approach than applying a given formula.

That said, I will leave you with a combinations/permutations question I encountered in real life…


A Quick Anecdote

My wife and I were set to board a plane. We knew we were in the same row, but my wife had forgotten whether the tickets were for adjacent seats. (There were very few seats left so my wife only remembered purchasing seats that were at least in the same row).

On this flight, the seats were arranged six in row (A – F), with the aisle splitting the seats into A-B-C and D-E-F. Knowing that I spend my day writing math problems, my wife asked me, “So, what is the probability we will be sitting next to each other?”

I thought about it for a moment, imagining us seated in different positions. I was about to begin the laborious process of mentally arranging us in all the different possible positions, but luckily stopped this brute force method and instead thought of the problem mathematically:

___ ___ ___      ___ ___ ___

My wife and I can be thought of as A and A. I then asked myself how many ways I could place A and A in the dashes above. Because A is redundant (as long as my wife and I are sitting next to each other, it doesn’t matter who is sitting on which side), I didn’t have to worry about order. I was essentially asking how many ways I could choose two dashes from six dashes. In formula-speak, this translates to 6C2 = 15.

Looking at the dashes above, I can only place A and A in four different positions, so that they are next to each other (remember, the space between the third and fourth dash is the aisle). Therefore, to find the probability of an event (my wife and I sitting next to each other), I divide the total number of possibilities (15 different ways that two people can sit next to each other) by the desired outcome – my wife and I seated next to each other: 4 possibilities.

I turned to my wife, eager to share my mathematical epiphany, “The chances of us sitting together was 4/15 or a little more than 25%. So, we are likely going to have to ask somebody to switch.”

Before I’d even gotten my last syllable out, my wife stopped me: “Actually, we are sitting together”. Holding both tickets in her hand, she added, “So much for your math.”


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37 Responses to The Difficulty of Context: Combinations and Permutations Questions

  1. Chandre August 13, 2014 at 10:21 am #


    In your practical example of your encounter while traveling you said, “I divide the total number of possibilities (15 different ways that two people can sit next to each other) by the desired outcome”. Is this not meant to be divide the total number of desired outcomes by the total number of possibilities? As this is the formula for probability

    • Chris Lele
      Chris Lele August 13, 2014 at 12:08 pm #

      Oops, I reversed things there. Thanks for catching that :). And yes, that is the formula for probability.

  2. Abhijeet Gautam August 17, 2013 at 10:13 am #

    Hi Chris,
    Can you please explain the answer to the fourth question?
    I am not really able to understand this question.


    • Chris Lele
      Chris Lele August 20, 2013 at 1:42 pm #

      Yeah, it’s a toughie :)!

      So, there are three different medications the doctor can prescribe, and the order of those medications is important. So we have to use the permutations formula.

      The trick to this problem is to remember that there are a total of seven days in which you can administer any three of these medications. So how many ways can you stagger the medicines over a week? Well, there are seven days for 3 medicines, so 7P3 = 210.

      Hope that helps!

      • Abhijeet Gautam August 23, 2013 at 10:08 pm #

        Hi Chris,
        Thanks for the explanation.

        I was flummoxed by the restriction : “If he can administer each drug once and cannot give more than one drug a day”.

        I broke the problem into stages, 7 of them, but that was incorrect. Only after your solution did I realize that there ought to be 3 stages, 1 for each medicine and that we got to fix days into these stages.

        Br, Abhijeet

        • Chris Lele
          Chris Lele August 26, 2013 at 11:46 am #

          Hi Abhijeet,

          Yeah, it’s a tricky question–but I think that’s exactly what the GRE is doing these days: making the setup of the question difficult, but the math relatively straightforward.

          • Meera October 2, 2014 at 8:46 am #

            The meds cannot be repeated in a week,
            So consider 3 stages,
            Stage 1. Case for P medicine , choose out of 7 days,
            Stage 2. Case for D med, choose out of 6 days
            Stage 3. Choose out of 5 days
            So total is 7 into 6 into 5 = 210 days

  3. Omair March 20, 2013 at 6:28 am #

    I wish I was in America so that i can take GRE classes live by magoosh faculty:(

    • Margarette Jung
      Margarette March 20, 2013 at 12:43 pm #

      Hi, Omair! We actually don’t offer in-person classes, so you’re not missing out on much, don’t worry :).

  4. sammy August 2, 2012 at 6:55 am #

    So, I don’t see any answers posted? Is the first one 35 and the second 210?

    • Chris Lele
      Chris August 2, 2012 at 5:15 pm #

      Yep, those answers are correct :).

  5. Ad July 22, 2012 at 10:33 am #

    Hi Chris
    Please help me solve this problem..I just cant figure out how to approach this:

    Right triangle PQR is to be constructed in the XY plane such that right angle is at P and PR is parallel to the X axis. The X and Y coordinates of P, Q and R are to be integers that satisfy the inequalities -5<=x<=4 and 7<=y<=17. How many different triangles with these properties could be constructed?
    (A) 110
    (B) 1100


    • Chris Lele
      Chris July 25, 2012 at 3:30 pm #

      This is a classic question from the GMAT Official Guide. THe good new is our GMAT guru Mike has made a video of this very problem of youtube:

      Hope that helps!

      • Ad July 26, 2012 at 12:29 am #

        Thanks a lot. The video was great! Can such questions of GMAT level be asked in the GRE?

        • Chris Lele
          Chris July 27, 2012 at 3:47 pm #

          No, I am pretty confident you will not get such a difficult question on the GRE.

          • Ad August 5, 2012 at 10:21 am #

            Hi Chris
            I would like to say a big thank you to Magoosh. I think all of you are doing a wonderful job. I just gave my GRE and got a combined score of 324 and I can safely say that most of it is due to my regular reading of the Magoosh blog. I cannot praise enough the wonderful job you all are doing. The posts, questions, explanations, advice- all of it is truly one of its kind and is the most helpful out of all the sources available in the market. Having given the GRE, I can safely say that the Magoosh advice and techniques helped me the most. My only regret is that I wish I would have come to know of Magoosh earlier and used its complete product instead of the hundreds of flimsy test prep books I actually used. Thank you once again for all your help.

            • Chris Lele
              Chris August 6, 2012 at 2:48 pm #

              Awesome! Congratulations on your amazing score :). We at Magoosh are very happy to have been part of it. Best of luck!

      • Akshay September 16, 2013 at 10:55 pm #

        Wow . Now that GMAT sum was a great math video ! Besides listening your lessons makes this topic sound easy . The medicine problem was a good one too . Got stumped actually despite understanding the concepts, typical of the GRE I suppose .keep up the great work

        • Chris Lele
          Chris Lele September 17, 2013 at 5:04 pm #

          Great! I’m happy I stumped you (in the sense that I’m getting you prepared for test day :)). Keep going back to all the permutations and combinations questions we have on the blog. The practice will make you better :).

  6. bebo June 29, 2012 at 2:23 am #

    Hi, could you please tell the solution to the second problem, students and seating arrangement. Thanks

    • Chris Lele
      Chris June 29, 2012 at 3:05 pm #


      Seven students can sit in chair 1, 6 in chair 2, and 5 in chair 3. So that gives us 7 x 6 x 5 = 210 (and no multiple students per chair :)).

  7. jay June 4, 2012 at 3:59 pm #

    What exactly in the second question tells us that the order is important –
    The information given in the bracket or the fact that there are three chairs?

    • Chris Lele
      Chris June 5, 2012 at 2:04 pm #

      Conceptually, it is the fact that any student can sit in one of three chairs. From this we can infer that each chair is different. Imagine the chairs in a row. If Mary is in the first chair, then she is in a different physical location than if she were in the second or third chair. We call this position. When the position matters, we want to use permutations.

      Hope that helps!

      • jay June 5, 2012 at 3:35 pm #

        Thanks chris that helped.

        I came across this question on probability. I wasn’t able to understand the solution. Could you please help.

        If there are 85 students in a statistics class and we assume that there are 365 days in a year, what is the probability that at least two students in the class have the same birthday (assuming birthdays are distributed independently)?

        A) (85/365) *(84/364)
        B) (1/365) *(1/364)
        C) 1- (85! / 365!)
        D) 1- {365! /(280! * 365^85)}

        • jay June 5, 2012 at 4:00 pm #

          This question is not from a book. A friend passed it to me so I’m not really sure wether these kind of questions actually come in the GRE or not.

          • Mithun April 24, 2016 at 11:32 pm #

            Answer is D

            • Mithun April 24, 2016 at 11:38 pm #

              The way 85 student can have different birthdays= 365P85

              total ways =365^85

              Probability of having same birthday for at least 2 student=

        • Chris Lele
          Chris June 5, 2012 at 4:00 pm #

          This is the classic probability problem for B-days. The math here is WAY beyond the scope of the GRE. So don’t worry about this question :).

          • jay June 5, 2012 at 4:03 pm #

            Alright. Thanks.

          • D December 7, 2014 at 6:27 am #

            Hi Chris, if you don’t mind, could you explain how you got the answer for No.3? (:

            • Chris Lele
              Chris Lele December 9, 2014 at 12:01 pm #


              A heptagon has seven sides. A triangle is made of three sides. That means three of any of the seven vertices make up a triangle. Another way of thinking about it is you can choose any three vertices and they will make a triangle. How many ways can you choose three from seven? 7C3 = 35.

              Hope that helps!

              • Ram January 7, 2015 at 9:49 am #

                Hi Chris,

                Isn’t the triangles that can be formed from a heptagon 35 instead of 210?


                • Jonathan January 7, 2015 at 4:12 pm #

                  Hey Ram!
                  You’re right -it’s 7C3 which is 35, not 210. We’ll fix this. Great catch! 🙂

  8. jay June 4, 2012 at 8:34 am #

    Hi chris

    I find permutations and combinations really hard. So i tried solving the initial questions, which others might find to be really easy. Are the answers for the first 4 questions in this blog 7C3,7C3,7C3 and 7P3?

    • Chris Lele
      Chris June 4, 2012 at 3:43 pm #

      Hi Jay,

      Permut./Comb. are indeed tricky! But you are definitely on the right track :). You only missed one of the questions. The second one, with students being arranged in chairs, is 7P3, because the order in which students are sitting is important. Meaning if I pick A, B, and C to sit in my three chairs, ABC is different from CBA: in the first case A is sitting on the left, in the second A is on the right.

      Let me know if that makes sense. If not, I’d be happy to explain in greater detail :).

  9. Aman April 14, 2012 at 5:28 am #

    Hi Chris,
    The example is EPITOME ,every time i get exhausted and bored of GRE studies your blogs makes me effusive.I always wonder how you manage to all these stuff ,may be your day is for 48 hours 😀 ,but now i got it .You have made GRE a part of your life .
    HATS OFF TO YOU….!!!

    As per this concept,what i found here (and in most P&C,Probability question ) is the order in which you arrange matters or not .
    It would be great if you can include more question like this .

    And the methods to solve the above mention four question ,need to confirm my choices.

    For helping us all, keep posting more

    • Chris Lele
      Chris April 16, 2012 at 11:42 am #

      I will indeed be posting more questions on whether a question requires combinations or permutations 🙂

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