It is common for GMAT students looking for a 700+ score to have many questions about GMAT combinatorics problems. These are the GMAT questions that ask you to count up all the possible arrangements of individuals and groups in a variety of situations: How many ways can 5 men and 5 women be ordered in a line? How many high fives occur in a group of 15 people on a basketball team? GMAT tutors often find themselves spending an inordinate amount of time helping students improve their ability to answer these types of GMAT questions.
We have previously written about how to solve GMAT combinatorics problems in previous blog articles on the principles of combinatorics. One previous article offered a variety of GMAT combinatorics practice problems.
The fundamental counting principle is at the heart of all GMAT combinatorics problems. This principle covers a critical rule governing the ways that items can be grouped, arranged, and combined. The fundamental counting principle states that if there are 2 ways that a certain thing can happen, and 4 possibilities of a different thing occurring after that, and 6 different ways a third event could unfold, then the total number of event possibilities is 2 * 4 * 6 = 48.
In other words, assuming events unfold independently, meaning one event occurring does not impact the chances that another occurs, you can multiply possibilities together to understand the total number of possible outcomes. If there are a ways this event can unfold, and b ways that event can unfold, the total number of outcomes is a x b.
However, for many life situations and GMAT combinatorics problems, events do not unfold independently. One event impacts another, particularly when you're dealing with arrangements. For example, let’s say you have 5 pairs of shoes, and you want to wear a different one each day of the school week. How many different ways can you decide to wear your shoes in this week? The first day, you have five options. The next day you have four (because you already wore one pair). On day three, you have three options. You have 5 x 4 x 3 x 2 x 1 ways to wear your shoes in this week, or 120 ways. We can use the factorial notation of 5! to communicate this.
Our next key principle for approaching GMAT combinatorics problems is the use the following rule of thumb: AND choices get multiplied, OR choices get added or subtracted.
Here’s a simple example. Jane is choosing an outfit for her first day at a new job. It consists of a skirt or a pair of pants, a shirt, a pair of shoes, and a jacket. She has 3 skirts, 5 pairs of pants, 7 blouses, 4 pairs of shoes, and 3 jackets. How many different outfits can she choose from?
In this example, the choice of whether to wear a skirt or a pair of pants is an exclusive choice – Jane can't wear both.
Jane will choose a pair of pants AND a blouse AND a pair of shoes AND a jacket OR she must choose a skirt AND a blouse AND a pair of shoes AND a jacket.
If she chooses a skirt, she has 3 x 7 x 4 x 3 = 252 outfits to wear. If she chooses pants, she has 5 x 7 x 4 x 3 = 420 choices. Note that these two sets of choices are mutually exclusive, so we can add them together and see that Jane has 672 outfit choices.
Here is an example that illustrates this principle.
Six candidates for student council are involved in a debate. There are six chairs for the students, but two of the students refuse to sit next to each other. How many possible arrangements are there for the students at the debate?
A) 120
B) 240
C) 480
D) 720
E) 840
Let's look at what we've got—there are 6 students, so if there were no restrictions at all, it would simply be 6! or 720 (be on the lookout for too-obvious answer choices like this one). You can also eliminate E, because the answer couldn’t be a number higher than 720.
Now, we know that we're talking about arrangements, so order matters.
Now for the restriction: two students refuse to sit next to each other. To keep it simple, let's just give our students letters, A through F. And let's say A and B won't sit together. Rather than trying to find all the ways in which A and B can be arranged, separated (which would take a while), let's just find all the ways that A and B can be seated next to each other, and subtract that from the total, which is 6! or 720. If we treat A and B as a single unit, there are 5! ways to arrange everybody, or 120.
AB C D E F
____ ____ ____ ____ ____
But there's one more step! Since order matters, AB is not the same arrangement as BA! So we need to multiply the 5! by all the possible arrangements of A and B (technically this is 2!, but it's pretty easy to see we've only got 2 options)
So the answer is 720 – 2(120) = 480.
To watch a free video tutorial on combinatorics, click here.