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Four couples are to select a group that consists of three [#permalink]
23 Jul 2008, 00:38

Four couples are to select a group that consists of three people from different couples. How many such groups are possible? (A) 24 (B) 28 (C) 32 (D) 36 (E) 40

P (selecting first person) = 8/8 ( can be any one) P (selecting second person) = 6/7 ( cant take spouse of first) P (selecting third person) = 4/6 (cant take spouses of first two)

for the first spot, we can pick any one person out of the 8, i.e. 8C1 = 8

for the second spot, we need to pick one person out of the remaining 6 ... not the remaining 7, since we dont want to pick the spouse of the person we selected first ... so 6C1=6

same logic for the third spot to get 4C1=4

finally, we have to divide out by 3!, since this represents the different permutations of the three spots, and we dont care about order so we end up with (8x6x4)/6 = 32