Keeping nine of twelve songs on an album

Question

A band recorded 12 songs and will keep 9 of them for an album (the running order is decided later, so right now only the set of songs matters). How many different sets of 9 songs can they keep?

Accepted Answer

This is \binom{12}{9}, choosing 9 from 12 with order ignored. The direct computation has large numbers, so use the symmetry shortcut: choosing 9 to keep is the same as choosing 3 to cut. Therefore \binom{12}{9} = \binom{12}{3} = \frac{12 	imes 11 	imes 10}{3 	imes 2 	imes 1} = \frac{1320}{6} = 220. This works because \binom{n}{k} = \binom{n}{\,n-k\,}: every "keep" set pairs with exactly one "cut" set. Takeaway. Use \binom{n}{k} = \binom{n}{n-k} to swap a large k for a small one. Here \binom{12}{9} = \binom{12}{3} = 220.

Keeping nine of twelve songs on an album

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