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## How many subsets of three elements each can be made from a set of five elements?

This is given by (5 choose 3) = 5! / [3!(5-3)!] = 5·4/2 = 10.

Here’s the idea behind the formula. Suppose you were to pick the elements one at a time. Then you would have 5 choices for your first element, four choices for your second element, and 3 choices for your third element. This is 5·4·3, which is the 5! / (5-3)! part of the formula.

However, when you’re picking a set, the order doesn’t matter; picking A,D,E is the same as picking D,E,A. Thus, once you have your 3 elements chosen, there are 3! = 3·2·1 = 6 ways of arranging them–again, there are 3 choices for the first element, 2 for the second, and only 1 for the third. Therefore, we need to divide by 3! in order to account for the 3! times we’ve included each set of 3. This is the factor of 3! in the denominator.

In general, if you have n objects, you have to choose m of them, and order doesn’t matter, then there are

(n choose m) = n! / [m!(n-m)!]

ways to do this.

(Obviously n must be larger than m for this to make any sense!)

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