{"id":422,"date":"2012-01-11T17:11:56","date_gmt":"2012-01-12T01:11:56","guid":{"rendered":"https:\/\/magoosh.com\/gmat\/?p=422"},"modified":"2020-01-15T10:51:02","modified_gmt":"2020-01-15T18:51:02","slug":"gmat-permutations-and-combinations","status":"publish","type":"post","link":"https:\/\/magoosh.com\/gmat\/gmat-permutations-and-combinations\/","title":{"rendered":"GMAT Permutations and Combinations"},"content":{"rendered":"

Permutations<\/h2>\n

A permutation<\/span> is a possible order in which to put a set of objects.\u00a0 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?\u00a0 Another way to say that is: 5 books have how many different permutations?<\/p>\n

In order to answer this question, we need an odd math symbol: the factorial.\u00a0 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.\u00a0 For example, 4! (read “four factorial”) is<\/p>\n

4! = (4)(3)(2)(1) = 24<\/p>\n

 <\/p>\n

Here’s the permutation formula:<\/p>\n

# of permutations of n objects = n!<\/p>\n

So, five books \"right\" the number of permutations is 5! = (5)(4)(3)(2)(1) = 120<\/p>\n

 <\/p>\n

Combinations<\/h2>\n

A combination<\/span> is a selection from a larger set.\u00a0 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?\u00a0 Another way to say that is: how many different combinations of 3 can be taken from a set of 20?<\/p>\n

This formula is scary looking, but really not bad at all.\u00a0 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<\/p>\n

# of combinations = \"{n!}\/{r!(n-r)!}\"<\/p>\n

 <\/p>\n

Again, this looks complicated, but it gets simple very fast.\u00a0 In the question just posed, n = 20, r = 3, and n \u2013 r = 17.\u00a0 Therefore,<\/p>\n

# of combinations =\u00a0\"{20!}\/{3!(17)!}\"<\/p>\n

 <\/p>\n

To simplify this, consider that:<\/p>\n

20! = (20)(19)(18)(17)(the product of all the numbers less than 17)<\/p>\n

 <\/p>\n

Or, in other words,<\/p>\n

20! = (20)(19)(18)(17!)<\/p>\n

 <\/p>\n

That neat little trick allow us to enormously simplify the combinations formula:<\/p>\n

# of combinations = \u00a0\"{(20)(19)(18)(17!)}\/{3!(17)!}={(20)(19)(18)}\/{3!}={(20)(19)(18)}\/{(3)(2)(1)}=1140\"<\/p>\n

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.<\/p>\n

Practice Questions<\/h2>\n

1) A bookseller has two display windows.\u00a0 She plans to display 4 new fiction books in the left window, and 3 new non-fiction books in the right window.\u00a0 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?<\/p>\n