# GRE Combinations and Permutations

Let’s go over a few definitions before we go over the different types of combination and permutation questions. The first thing to know is what “!” means. In general, you need to remember the formula for combination and permutation which involves the ! sign.

“!” is read aloud as “factorial”. For any integer n, n! = n*(n–1 )*(n–2)*…*3*2*1.

For example, 6! = 6*5*4*3*2*1

### Combinations

Combinations are used when the order of the objects doesn’t matter. The keyword that lets you know to use the combination formula is the word **choose**. When you choose something, it doesn’t matter what order you choose them in.

The general formula for choose **r** objects out of **n** objects (where r is obviously less than or equal to n) is:

* n*! / **( n – r)! r!**

**Example:**

I am trying to create a custom striped tie and I need to pick two colors for it out of 7, then how many color combinations can I have. The answer is to choose 2 colors out of 7.

In this case, n = 7, r = 2 and the answer is 7! / (5! * 2!) = **21**

### Combinations of Combinations

What if the question tells you that you have a number of choices for one thing, and another number of choices for another, how many combinations can you have of the two things?

**Example:**

Any buyer of a new sports car has to pick between 2 of 5 options for seat colors and 3 of 4 options for dashboard accessories. How many different combinations of colors and dashboard options are available to this buyer?

We look at seat colors first. Choosing 2 out of 5 and applying the formula, we get 10. Then choosing 3 accessories out of 4, we get 4 combinations. So in total, we have 10*4 seat color-accessories combinations, because for every seat color combination, there are 4 accessory combinations that we could match with it. With 10 seat-color combinations, we thus have 40 accessory combinations to match with it.

### Permutations

With permutations, order matters.

**Example:**

If we have 5 people and five seats on a plane, there are 120 ways of seating these 5 people. If all the seats are empty and I want to fill the first seat, I have 5 people to choose from to fill the seat. Then for the second seat, I have 4 people to choose from (because one person has already sat down). And for the third seat, I have 3 people and so on. So the answer is 5!

If I have 7 people and five seats, the same concept applies – 7 * 6 * 5 * 4 * 3 where each number represents the number of people I can choose from to put in each seat starting from the left.

### Permutations in a Circle

What if you need to sit people around a circular table? Since there is no longer a left end and a right end, the first person you seat is just a point of reference.

**Example: **

If you are seating 5 people in 5 seats around a circular table, instead of having 120 seating arrangements like before, you have 24 arrangements.

In general, when arranging n people in a circle, there are (n-1)! Number of ways.

### Permutations with Repeated Terms

This type of question usually asks, how many different words can you form from the letters of word. And more often than not, the word will have repeated letters in it.

**Example: **

How many 7-letter words can you form from the word APPEASE?

There are 7 letters so there are 7! ways of arranging these letters in order. But P is repeated twice, and since one P looks the same as the other P, you have to divide 7! by 2!. A and E are also repeated twice so you have to divide by 2! twice more.

The final answer is

Practice Combinations and Permutations at grockit.com/gre or get more info by following us @grockit!

nice but this information is not sufficient. give the information about digit problems in this.

What about problems that ask how many ways the letters of a word can be arranged if all the vowels have to sit next to one another? I still cannot figure that one out.

Lauren, obviously in a word all vowels wont sit all next to one another.But if some letters need to be next to each other you could just use their combination as one letter and play with (n – the number of letters lost by grouping) letters.. (then you should also multiply by the factorial of the number of letters in each group so that take all cases :p )

I have this question:

A bank generates a 4-digit pin code for each customer. What is the probability that a random pin will not contain any repeated digits?

This is combination and probability right?

Good stuff. Thanks!