By: Rich Zwelling, Apex GMAT Instructor
Date: 4th March, 2021

In our previous post, we discussed how GMAT combinatorics problems can involve subtracting out restrictions. However, we discussed only PERMUTATIONS and not COMBINATIONS.

Today, we’ll take a look at how the same technique can be applied to COMBINATION problems. This may be a bit more complicated, as you’ll have to use the formula for combinations, but the approach will be the same.

Let’s start with a basic example. Suppose I were to give you the following problem:

The board of a large oil company is tasked with selecting a committee of three people to head a certain project for the following year. It has a list of ten applicants to choose from. How many potential committees are possible?

This is a straightforward combination problem. (And we know it’s a COMBINATION situation because we do not care about the order in which the three people appear. Even if we shift the order, the same three people will still comprise the same committee.)

We would simply use the combination math discussed in our Intro to Combination Math post:

 10C3 =       ————-
                     3! (10-3!)


3! (7!)





= 120 Combinations 

However, what if we shifted the problem slightly to look like the following? (As always, give the problem a shot before reading on…):

The board of a large oil company is tasked with selecting a committee of three people to head a certain project for the following year. It has a list of ten applicants to choose from, three of whom are women and the remainder of whom are men. How many potential committees are possible if the committee must contain at least one woman?

A) 60
B) 75
C) 85
D) 90
E) 95

In this case, there’s a very important SIGNAL. The language “at least one” is a huge giveaway. This means there could be 1 woman, 2 women, or 3 women which means we would have to examine three separate cases. That’s a lot of busy work. 

But as we discussed in the previous post, why not instead look at what we don’t want and subtract it from the total? In this case, that would be the case of 0 women. Then, we could subtract that from the total number of combinations without restrictions. This would leave behind the cases we do want (i.e. all the cases involving at least one woman). 

We already discussed what happens without restrictions: There are 10 people to choose from, and we’re selecting a subgroup of 3 people, leading to 10C3  or 120 combinations possible. 

But how do we consider the combinations we don’t want? Well, we want to eliminate every combination that involves 0 women. In other words, we want to eliminate every possible committee of three people that involves all men. So how do we find that?

Well, there are seven men to choose from, and since we are choosing a subgroup of 3, we can simply use 7C3 to find the number of committees involving all men:

7C3 =       ————-
                 3! (7-3!)


3! (4!)



= 7*5 = 35 Combinations involving all men

So, out of the 120 committees available, 35 of them involve all men. That means 120-35 = 85 involve at least one woman. The correct answer is C. 

Next time, we’ll return to probability and talk about how the principle of subtracting out elements that we don’t want can aid us on certain questions. Then we’ll dovetail the two and talk about how probability and combinatorics can show up simultaneously on certain questions.

Permutations and Combinations Intro
A Continuation of Permutation Math
An Intro To Combination Math
Permutations With Repeat Elements
Permutations With Restrictions
Combinations with Restrictions
Independent vs Dependent Probability
GMAT Probability Math – The Undesired Approach
GMAT Probability Meets Combinatorics: One Problem, Two Approaches