Do you know the difference between permutation and combination? No? You’re not alone. When it comes to GRE data analysis, combinations and permutations are the bane of many students. Yet, what I’ve noticed over the years is it’s not so much both of them that are the issue as it is which one to use for a particular problem: the combination vs permutation question. In other words, students have no difficult identifying whether a question is a combinations/permutations problem. The difficulty is in knowing exactly which one it is—combinations or permutations?

## Combinations and Permutations

One way to think of it is to think of **permutations** as the number of arrangements or orderings within a fixed group. For example, if I have five students and I want to figure out how many ways they can sit in five chairs, I’m going to use the permutations formula. First off, the number in the group is fixed. Secondly, I’m looking for how many ways I can “arrange” the students in five chairs.

**Combinations**, on the other hand, are useful when figuring out how many groups I can form from a larger number of people. For instance, if I’m a basketball coach and I want to find out how many distinct teams I can form based on a group of people, I want to use combinations.

To make sure you understand this important distinction, here are three different scenarios. Your job is not to solve the question but to determine whether you use the combinations or the permutations formula to solve them.

## Combination vs Permutation: You Decide

1. Joan has five panels at home that she wants to paint. She has five different colored paints and intends to paint each panel a different color. How many different ways can she paint the five panels?

2. How many unique combinations of the word MAGOOSH can I form by scrambling the letters?

3. There are seven astronauts who are trying out to be part of three-person in-space flight team. How many different flight teams can be formed?

## Difference Between Combination and Permutation: Answers

1. Permutations

She wants to arrange the colors. The number of panels is fixed. Had she been choosing five panels from a total of 8, let’s say, then we would need to use combinations.

2. Permutations

Okay, this was a little bit of a trick, since I used the word “combinations”. But that word I used colloquially, not mathematically. In this case, the number of letters is fixed. We are simply rearranging them.

3. Combinations

We are choosing a group from a larger group. One way to think of it is that when you use the word choose in the context of selecting from a group, you are dealing with combinations. And “choose” and “combinations” both begin with the letter ‘C’. There is an exception to this rule that I’ll talk about in the next section.

## Difference Between Permutation and Combination: The Main Point

Big Idea: If you are forming a group from a larger group and the placement within the smaller group is important, then you want to use permutations.

Imagine a group of 12 sprinters is competing for the gold medal. During the award ceremony, a gold medal, silver medal and bronze medal will be awarded. How many different ways can these three medals be handed out?

Remember that with permutations ordering is key. Even though the top three spots for the sprinters forms a subgroup, it is the ordering within that subgroup that matters greatly, and is the difference between a gold, silver, and a bronze medal. An easy way to solve this question mathematically is to imagine that the dashes below are the podium upon which each sprinter will stand (albeit the dashes are at the same level):

____ ____ ____

gold silver bronze

To find out the number of different arrangement, ask yourself how many athletes can stand on the gold medal podium? Well, we have a total of 12 athletes. What about the silver medal podium? Now, we have one athlete fewer—since one is already on the gold medal podium. So that gives us a total of 11 for the silver medal spot. Finally, that leaves us with 10 athletes for the bronze medal podium.

The math looks like this:

12 x 11 x 10 = 1320

You might notice that this is the fundamental counting principle. The idea that when we are looking for total number of outcomes, we multiply numbers—or in this case, the number that goes on top of each dash—together. For instance, if I have six pairs of shorts and four pairs of t-shirts and I’m wondering how many different combinations of shorts and t-shirts I can wear, I want to multiply each, not multiply them:

___4_______ x _____6______ = 24

# of shirts # of shorts

I do not want to add them, which would give me 10, the wrong answer, in this case.

The reason I bring up the fundamental counting principle is that some questions will actually combine combinations with the fundamental counting principle (though you’ll likely use the fundamental counting principle more often with permutation questions). To give you an idea of how a combinations can show up along with the fundamental counting principle, try the following question:

Mrs. Pearson has 4 boys and 5 girls in her class. She is to choose 2 boys and 2 girls to serve on her grading committee. If one girl and one boy leave before she can make a selection, then how many unique committees can result from the information above?

(A) 9

(B) 12

(C) 18

(D) 22

(E) 120

The first step in this problem is recognizing whether we are dealing with combinations or permutations. Since, I am ‘choosing’ from a larger group, in this case two separate groups, I want to use combinations. Remember: once we’ve chosen the 2 boys or 3 girls, the position within the committee doesn’t matter. That is, either you are in the committee or out of the committee (there are no gold medalists here!)

The next thing to notice with this problem is that of the original 9 students, 2 leave, one boy and one girl. So that leaves us with 3 boys and 4 girls. We want to choose two each. Therefore, we have to set up one combination for boys and one for girls.

For boys, we have 3C2 and for girls we have 4C2. This gives us: 3C2 = 3 and 4C2 = 6

The final step is once we’ve figured out the combinations above, we have to use the fundamental counting principle and multiply the total number of possibilities in a committee, not add them: 3 x 6 = 18, answer (C).

Many students get stuck on this step and wonder, why don’t I add them. It’s a good question, so what I want you to do is imagine that you have 3 shirts and 3 pants. How many different shirt-pants getups can you wear? Well, for each shirt there are 3 options of pants. Therefore, we multiply and get 9.

My advice is to try about 5 or 6 more combinations/permutations problems so that you can get the hang of it. With a little practice, you’ll able to deal with most of the problems the GRE hands you. Even if you miss a question—likely because it is very difficult—the fundamentals in this post should be enough to help you understand the explanation to that question, so that you can get a similar question right in the future.

*Editor’s Note: This post was originally published in May 2011 and has been updated for freshness, accuracy, and comprehensiveness.*