## 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 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 =

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 =

To simplify this, consider that:

20! = (20)(19)(18)(17)(the product of all the numbers less than 17)

Or, in other words,

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

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

# of combinations =

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

## Practice Questions

1) 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?

- 24
- 72
- 144
- 336
- 420

2) 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?

- 15
- 30
- 90
- 150
- 360

## Answers and Explanations

1) 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.**

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

# of combinations = =

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, & 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.**

hi,

Could you please provide some more practice questions based on the above concept. Thanks!

Dear Shilpi

Take a look at this post:

http://magoosh.com/gmat/2013/difficult-gmat-counting-problems/

Mike

Thanks a lot

Dear Shilpi,

You are quite welcome. Best of luck to you!

Mike

“With the plays, we are taken a combination of 3 from a set of 5″

Should be 2 from a set of 5. Otherwise, a very good post.

Steven,

Thanks for catching that — I just fixed it. Incidentally, I probably made the mistake because, as you may know, 5C2 = 5C3 — at a certain level of analysis, there’s no difference picking two from five or three from five. But for clarity, it’s good to make this change. Thanks,

Mike

Hi Mike

Can we do it this way?

1/3 * 2/5 * 1/4 = 1/30

so we have 30 combination?

Thanks

Naren,

I assume you mean practice question #2. Yes, that would be another way to compute the number of possible reading lists.

Mike

Mike – Thanks for the valuable information. I have a clarification to make with respect to question 1.

I was thinking – would not the answer be 24 + 6 = 30. I am thiking as she wants the ‘ficiton’ books still too be placed on the left hand side in any order and the same goes for ‘non fiction books.

Am i missing a trick here ?

Rahul Sehgal

GMAT aspirant !!

Dear Rahul,

The big conceptual piece you are missing is the Fundamental Counting Principle. See this blogpost:

http://magoosh.com/gmat/2012/gmat-quant-how-to-count/

Let me know if that doesn’t clear up why we *multiply* instead of adding.

Mike

A nice intro to P&C and, probably you are true WRT the GMAT, this is probably more than enough to crack

Thank you very much.

Mike