If a committee of 3 people is to be selected from among 6 married couples such that the committee does not include two people who are married to each other, how many such committees are possible?
(A) 20
(B) 40
(C) 80
(D) 120
(E) 160
Official Solution
Credit: Veritas Prep
We have 6 married couples i.e. we have 12 people. Out of these 12 people, we have to choose 3. If there were no constraints on the choice and we could choose any 3 out of these 12, in how many ways could we do that?
12 people and 3 available positions – We use basic counting principle to arrive at 12*11*10. But, recall that we have arranged the 3 people here. To un-arrange, we divide this by 3! to get 12*11*10/3! arrangements. This is what we discussed last week.
Moving forward, this question has a constraint – no two people married to each other should be selected i.e. at most one of any two married people could be selected. Say, if we select Mr. X, we cannot select Ms. X and vice versa.
Let’s try to use our basic counting principle and un-arranging method here:
You have 12 people and 3 positions. In the first position, you can put any one of the 12 people. In the second position, you can put any one of 10 people (not 11 because the spouse of the person put in 1st position cannot be put in the second position). In the third position you can put any one of 8 people. (We have already selected 2 people for the previous two places and their spouses cannot be selected for the third position so 4 of the 12 people are out.)
Total number of arrangements = 12*10*8
But mind you, these are arrangements. We just need a group of 3 people. They don’t need to be arranged in the group. So we divide these arrangements by 3! to just ‘select’ the people.
Number of committees possible = 12*10*8/3! = 160
Bunuel Could you please update the original Q with the answer choices? Thanks!