# Permutations and Combinations

## Permutations

A permutation is a possible order in which to put a set of objects. Suppose I had a shelf of 5 different books, and I wanted to know: in how many different orders can I put these 5 books? Another way to say that is: 5 books have how many different permutations?

In order to answer this question, we need an odd math symbol: the factorial. It’s written as an exclamation sign, and it means: the product of that number and all the positive integers below it, down to 1. For example, 4! (read “four factorial“) is

4! = (4)(3)(2)(1) = 24

Here’s the permutation formula:

# of permutations of n objects = n!

So, five books right the number of permutations is 5! = (5)(4)(3)(2)(1) = 120

## Combinations

A combination is a selection from a larger set. Suppose there is a class of 20, and we are going to pick a team of three people at random, and we want to know: how many different possible three-person teams could we pick? Another way to say that is: how many different combinations of 3 can be taken from a set of 20?

This formula is scary looking, but really not bad at all. If n is the size of the larger collection, and r is the number of elements that will be selected, then the number of combinations is given by

# of combinations = \dfrac{n!}{r!(n-r)!}

Again, this looks complicated, but it gets simple very fast. In the question just posed, n = 20, r = 3, and n - r = 17. Therefore,

# of combinations = \dfrac{20!}{3!(17)!}

To simplify this, consider that:

20! = (20)(19)(18)(17)...(2)(1)

Or, in other words,

20! = (20)(19)(18)(17!)

That neat little trick allow us to enormously simplify the combinations formula:

# of combinations = \dfrac{(20)(19)(18)(17!)}{3!(17)!} = \dfrac{(20)(19)(18)}{3!} = \dfrac{(20)(19)(18)}{(3)(2)(1)} =1140

That example is most likely harder than anything you’ll see on the QUANT math, but you may be asked to find combinations with smaller numbers.

## Practice Questions

Q1. A bookseller has two display windows. She plans to display 4 new fiction books in the left window, and 3 new non-fiction books in the right window. Assuming she can put the four fiction books in any order, and separately, the three non-fiction books in any order, how many total configurations will there be for the two display windows?

(A) 24

(B) 72

(C) 144

(D) 336

(E) 420

Q2. The county-mandated guidelines at a certain community college specify that for the introductory English class, the professor may choose one of three specified novels, and choose two from a list of 5 specified plays. Thus, the reading list for this introductory class is guaranteed to have one novel and two plays. How many different reading lists could a professor create within these parameters?

(A) 15

(B) 30

(C) 90

(D) 150

(E) 360

Q1. The left window will have permutations of the 4 fiction books, so the number of possibilities for that window is

permutations = 4! = (4)(3)(2)(1) = 24

The right window will have permutations of the 3 non-fiction books, so the number of possibilities for that window is

permutations = 3! = (3)(2)(1) = 6

Any of the 24 displays of the left window could be combined with any of the 6 displays of the right window, so the total number of configurations is 24*6 = 144