An Easier Approach to GRE Combination and Permutation Problems
Does seeing a question like this fill you with dread and anxiety? If you’re like most GRE students, there’s a good chance that it does. Questions that ask you to determine the number of groups that can be selected from a larger group, or that ask you to determine the number of ways a smaller group of entities can be arranged, are called GRE Combination and Permutation problems. If you’ve been studying for the GRE, you know these problems. For many students, these questions seem extremely difficult, if not impossible.
Well, I have good news: these questions aren’t actually all that difficult. What’s difficult, for many students, is memorizing large, convoluted formulas like this:
Now, formulas are certainly valuable, and they serve an important service: by memorizing the formula, you’ll always have a shot at answering these questions correctly. But if you have difficulty remembering the precise way to set up the formulas in these Combinations and Permutations questions, then rest assured: there is another way.
An Alternate Approach
Let’s start with this simple question: how many ways can seven people stand in a line? In this question, you’re asked for the number of arrangements that is possible when there are seven people. For those of you familiar with the mathematical term “factorial,” you know that 7 factorial, or 7!, is the best way to represent this value. Factorial means multiplying a number by itself and every integer smaller than that number, all the way down to 1. So 7! is the same as 7x6x5x4x3x2x1.
Ok, that’s great to know, but what’s even better to know is why 7! gives us the number of possible arrangements of a group of seven people.
To determine that value, imagine that we are arranging those seven people by placing each of them in one of seven spots in a row, like this:
__ __ __ __ __ __ __
How many people could go in that first spot? Seven, of course. Now, once we’ve placed one of the seven people in that first spot, how many people are left for the second spot? There are six people left, so that means there are 7×6 ways to pick two people and arrange them when there are seven people to choose from. What about the third person in the row? Well, two have already been assigned spots, so there are only five to choose from. So there are 7x6x5 ways to put three people in order when selecting from seven people. You see the pattern. We can go all the way down the line until that last spot is filled. That’s why the number of possible arrangements of seven people is 7!.
“Now, that’s all well and good,” you may be saying to yourself, “but how does that help me answer Combination and Permutation questions on the GRE?” Ah, but see, we were already figuring them out in the paragraph above!
Let’s start with Permutation questions. These ask you to determine the number of possible arrangement s of a group of entities. Whenever order matters, you know you’re dealing with a Permutation question. In the example above, we said that there were seven people to choose for the first spot, then six to choose for the second spot, and five to choose for the third spot. So 7x6x5 represents the number of possible ways to arrange three people, when there are seven people to choose from.
If you are asked to select a smaller group of entities from a larger group, and then put those entities in order, simply follow the same pattern you see above. Start with the number of total entities you have, then count down the number of entities you’re selecting, multiplying the values as you go.
So, if you have 11 total entities and you want to figure out how many ways you can select and order 4 of those entities, just count down from 11 until you have four total numbers: 11x10x9x8. That’s it!
So that’s how you can handle Permutations problems easily.
The other question type, called Combination, is very similar but there is one big difference: these questions simply ask you for the number of possible groups that can be picked from a larger selection. In other words, order doesn’t matter here. If I pick Bob, Mary, and Jim for my trivia team, that is one group. It doesn’t matter if it’s Mary, Jim, and Bob or Jim, Bob, and Mary. It’s always the same group.
To tackle these types of questions, we’ll utilize the same method as above, but we’ll add one extra step. For each group that we select, we’ll need to remove all of the permutations, or arrangements, that exist within that group. To do that, just divide the number of total permutations by the number of permutations that would exist in each group.
To demonstrate, let’s go back to the same example above, selecting 4 entities from a larger group of 11. We said already that the number of arrangements, or permutations, would be 11x10x9x8. To remove the permutations that exist within each group of 4, divide that entire amount by 4!, or 4x3x2x1. As a fraction, it looks like this:
The 4 and the 2 cancel out the 8, and the 3 divides into the 9. What remains in the numerator is 11x10x3, so there are 33×10, or 330 possible groups of 4 from a pool of 11 entities.
No calculator, no formulas, no worries. I told you these problems weren’t that tough!