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Ryan and 4 friends want to play a game. If only 3 people may play at a time, how many different combinations of 3 players are possible?

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1 Answer

Since it's Ryan and 4 friends, we have a total of 5 people wanting to play. Only three can play at a time. Since order is not important, it would be a combination instead of a permutation:
C(n,k) = n!/[(n-k)! k!]

C(5,3) = 5!/[(5 - 3)! 3!]

5! can be rewritten as (5)(4)(3!) (implied multiplication)

C(5,3) = [(5)(4)(3!)] / [(2!)(3!)]

Since a 3! exists in both the numerator and denominator, we can reduce:
C(5,3) = [(5)(4)]/[2!]

Since 2! = 2,
C(5,3) = [(5)(4)]/[2],
we can reduce the numerator and denominator to get
C(5,3) = (5)(2)/1 = 10

So there are 10 different combinations of 3 players. Brute force check: Assume the players are A, B, C, D, and E

1 - A, B, C
2 - A, B, D
3 - A, B, E
4 - B, C, D
5 - B, C, E
6 - C, D, E
7 - A, C, D
8 - A, C, E
9 - B, D, E
10 - A, D, E

Since order is not important #1 is the same as B, A, C; C, A, B; and B, C, A