GRE Permutations and Combinations

Updated March 04, 2010

The GRE Quantitative Section contains all sorts of Problem Solving questions. Some of them involve Permutations and Combinations

What You Need to Know

There’s not a lot of permutations and combinations involved in the GRE, so relax. Out of the 28 math questions, you’re unlikely to get more than 1 or at most 2 permutation questions.

That said, if you want to make sure to get the top GRE score, it’s always worth going over the basics. So here’s what you need to know:

Factorials

This basic idea is pretty simple. Here’s the mathematical expression: n! = n(n - 1)(n - 2) ... (2)(1) . For example: 10! = 10 * 9 * 8... * 2 * 1 . Factorials are used to order n elements- for example, say you want to arrange 5 kids in a row. There are 5! or 120 ways to do so.

More arrangements:

Say there are 5 pizza toppings, and you want to choose 2. How many options for pizza do you have? Well, you can count them- AB, AC, AD, AE, BC… that takes too long. You can use the permutation/combination idea instead. For the first topping, you have 5 options. For the second, since you already chose 1 topping, you now have 4 options. But, since it doesn’t matter whether you choose A first and B second or the other way around, you need to divide this by 2!. The answer is $\frac{5 * 4}{2!}= 10$ .

The idea is: Figure out how many spots you need to fill, then how many options you have per spot. If the order doesn’t matter, divide it out (using the factorial). That’s it.

Examples:

I want to choose 4 courses to study next year, and their order does matter. I have 9 courses to choose from. How many ways can I choose my courses?

Answer: For the first course, I have 9 options. For the second, 8, and so on. Remember that since order matters, you don’t divide out any factorials. Thus the answer will be 9 * 8 * 7 * 6 = 3024 .

I have 5 different colors.
Column A: The number of ways to paint 2 balls in different colors.
Column B:The number of ways to paint 3 balls in different colors.

Answer: Here, the order doesn’t matter, since all the balls are implied to be the same. So, for column A, we have 5 options for ball 1, 4 for ball 2, but we need to divide by the order of 2!, so we have $\frac{5 * 4}{2!} = 10$ . For column B we end up with $\frac{5 * 4 * 3}{3!} = 10$ too, so the columns are equal, the answer is C.