How to Solve Combinatorics Questions on the GMAT

Of all the topics on the GMAT quant section, few get students as confused as the concept of combinatorics. The concept of going to the store and picking up one of four possible gifts for a niece is pretty straightforward (she generally likes Barbies© or My Little Pony© toys), but picking up two toys out of four for your twin nieces and then deciding which one gets which often deteriorates into an exercise of brute force combinatorics.

Most people suspect that there’s a simple way to calculate the possible combinations, but they don’t bother to learn it because you can always figure it out with a pen and paper (or an abacus) and copious amounts of spare time.

But what happens when the exam asks you a combinatorics question that will take too long to solve via brute force? Well, figuring out what the question is asking is the first step towards success. The two first issues you should always address are: Does order matter, and is repetition allowed? If the order of the elements matters, you’re looking for permutations, not combinations. The difference is that you can get Barbie for Erin and a pony for Anne or Barbie for Anne and Barbie for Erin in this scenario, whereas if you’re buying two gifts for the same child then order is moot. Repetition is not allowed in this scenario because you won’t get her the same gift twice (or if you do you’re likely headed to the nursing home).

The general rule for permutations is that, if you’re looking for k toys out of n possible options, your options are defined by n!/(n-k)! (“!” Denotes factorial, which is taking the product of that integer and all integers smaller than it). In the example above, you have n=4 and k=2, so (4*3*2*1) / 2*1 options. It becomes quickly obvious that writing down the *1 at the end of every permutation adds nothing, so really this is (4*3*2)/2, which we can simplify to 4*3 or 12. If the order mattered you would have had fewer options, so determining the importance of order is paramount to getting the right answer.

Now that we’ve jogged our memories about combinatorics, let’s review a problem where understanding is as important (no, more important… ok, as important) as knowing the formula.

How many 5 digit numbers can be formed which are divisible by 3 using the numerals 0, 1, 2, 3, 4, 5 without repetition.

A)     120

B)      216

C)      240

D)     720

E)      3152

Blindly applying the formula we just discussed would yield n=6 and k = 5, so 6!/(6-5)! which is just 6! and therefore 720. Answer choice D is waiting for us if we go down this route, and it seems like a safe guess if you have no idea how to proceed, so a fair number of students may be tempted by this option. However, the formula gives you all possible permutations of these numbers, including, for example, 01234, which does not meet the second criterion of the question: That the numbers be divisible by 3. This restriction will eliminate many of the possibilities, so 720 possibilities (and 3152 even more so) is overshooting the target.

So what do we do? Go through all 720 possibilities and whittle the list down for every possibility that isn’t divisible by 3? That’s a good way to spend an unproductive afternoon! Why don’t we try and limit the number of possibilities by seeing which combinations of digits give us something divisible by 3 and then going through the permutations for those digits only? The rule for divisibility by 3 is that the sum of the digits must be divisible by 3. The fact that each digit must be unique means we only have six possibilities of digits, each possibility ignoring one of the six digits provided.

Digits 0,1,2,3,4 only: sum 10. Does not work.

Digits 0,1,2,3,5 only: sum 11. Does not work.

Digits 0,1,2,4,5 only: sum 12. Works!

Digits 0,1,3,4,5 only: sum 13. Does not work.

Digits 0,2,3,4,5 only: sum 14. Does not work.

Digits 1,2,3,4,5 only: sum 15. Works!

So we only really need to calculate the permutations of {1,2,3,4,5} and {0,1,2,4,5}. The first one is easier, as we can just take 5!/0! (0! Is 1 by definition) which gives us 120 possibilities. The second one contains a famous GMAT trick, which is that it cannot begin by 0 and still be a 5-digit number. Just like 7 is a one-digit number and not a five-digit number by adding an arbitrary number of 0’s at the beginning of it (James Bond’s understudy 00007). This logic dictates that the second set has only four possible numbers in the first position, and then 4 again for the second (any of the four that wasn’t put in the first spot) and then three, two, one as normal. Instead of the standard 5! we get 4 * 4! which is 96.  As such, the correct answer is 120 + 96 = 216. Answer choice B.

As we can see from the answer choices provided, the GMAT test makers know the mistakes you’re likely to make and prepare answer choices to trap you in case you make a simple mistake or absent-minded omission. Answer choices A (5!), C (2*5!) and D (6!) are all traps designed to make you feel like you got the right answer even though you made a mistake. Be particularly wary of answer choice C, most people who pick this answer feel confident that they got the question right. If you recognize that understanding the parameters of the question is the first step to getting the correct answer, you’ll be ready for any permutation of questions the GMAT throws at you..

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Ron Awad is a GMAT instructor for Veritas Prep based in Montreal, bringing you weekly advice for success on your exam.  After graduating from McGill and receiving his MBA from Concordia, Ron started teaching GMAT prep and his Veritas Prep students have given him rave reviews ever since.

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