Manjareev wrote:A committee of 3 people is to be chosen from four married couples.
what is the number of different committees that can be chosen if two people who are
married to each other cannot both serve on the committee?
16
24
26
30
32
We are given that there are four married couples (or 8 people) and we need to determine the number of ways of choosing 3 people in which no two people are a married couple. This is a special combination problem. Before we tackle this problem, let's review a combination problem with no restrictions.
With no restrictions, the number of ways of choosing 3 people from 8 people is 8C3, which is calculated as follows:
8C3 = 8!/[3!(8-3)!] = 8!/[3!5!] = (8 x 7 x 6)/3! = 56
8, 7, and 6, in the numerator, represent the number of ways the first, second, and third person can be chosen, respectively. We divide the numerator by 3! because in a combination problem we do not care about the order in which the 3 people are chosen.
However, in this (special combination) problem, 3 people are chosen in which no married couple can serve together on the committee. The first person could be any one of the 8 people. However, once a person is selected, that person's spouse cannot also be selected for the committee. This reduces the choices for the second person to 6 possible people (one person has already been selected and that person's spouse now cannot be selected). Once the second person is chosen for the committee, that person's spouse cannot be chosen. This reduces the number of people who could be chosen as the third person to 4. Therefore, the number of ways of choosing these 3 people is:
(8 x 6 x 4)/3! = 32
Thus, there are 32 ways to choose such a committee.
Answer:
E
