Question: From a group of 4 married couples, a team of 3 to be selected such that only one person from a couple is selected.

A) 16
b) 24
c) 26
d) 30
e) 32


Let's consider the case where couple #1 is on the committee. The third person can be chosen from any of the remaining six. So there are 6C1 = 6 ways of choosing the committee where both members of couple 1 are on the committee. The same reasoning holds for couples 2,3 and 4. So there are 6*4 = 24 "bad" ways of choosing the committee.

If there are 8C3 = 56 total ways of choosing the committee, and 24 "bad" ways of choosing the committee where a couple is on the committee, then that leaves us with 32 ways of choosing the committee where both members of the committee are NOT on the board.