Choosing a three-person committee from nine

Question

A club has 9 members and wants to form a committee of 3 people. The committee has no titles or roles, so only the group of people matters, not any ordering. How many different committees are possible?

Accepted Answer

Because order does not matter, this is a combination: "choose 3 from 9," written \binom{9}{3}. One clean way to see the formula: there are 9 	imes 8 	imes 7 ways to pick an ordered list of 3 people, but each unordered committee of 3 can be listed in 3! = 6 orders. Dividing removes that overcount: \binom{9}{3} = \frac{9 	imes 8 	imes 7}{3 	imes 2 	imes 1} = \frac{504}{6} = 84. Takeaway. When you select a group and order does not matter, use \binom{n}{k} = \dfrac{n!}{k!\,(n-k)!}. Choosing 3 of 9 gives 84.

Choosing a three-person committee from nine

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