# 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

## Answers and Explanations

**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

**Answer: C**

**Q2**. There are three possibilities for the novel. With the plays, we are taken a combination of 2 from a set of 5 right n = 5, r = 2, n - r = 3

# of combinations = \dfrac{5!}{2!3!} = \dfrac{(5)(4)(3)(2)(1)}{(2)(1)(3)(2)(1)} = \dfrac{(5)(4)}{2} = 10

If the plays are P, Q, R, S, and T, then the 10 sets of two are PQ, PR, PS, PT, QR, QS, QT, RS, RT, and ST.

Any of the three novels can be grouped with any of the 10 possible pairs of plays, for a total of 30 possible reading lists.

Answer: B