A committee of three people is to be chosen from four married couples. What is the number of different committees that

Gmat_mission wrote: ↑

Wed Mar 24, 2021 7:17 am

A committee of three 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?

A. 16
B. 24
C. 26
D. 30
E. 32

Answer: E

Source: GMAT Prep

Take the task of selecting the 3 committee members and break it into stages.

Stage 1: Select the 3 couples from which we will select 1 spouse each.
There are 4 couples, and we must select 3 of them. Since the order in which we select the 3 couples does not matter, we can use COMBINATIONS
We can select 3 couples from 4 couples in 4C3 ways (4 ways)

Stage 2: Take one of the 3 selected couples and choose 1 person to be on the committee.
There are 2 people in the couple, so this stage can be accomplished in 2 ways.

Stage 3: Take one of the 3 selected couples and choose 1 person to be on the committee.
There are 2 people in the couple, so this stage can be accomplished in 2 ways.

Stage 4: Take one of the 3 selected couples and choose 1 person to be on the committee.
There are 2 people in the couple, so this stage can be accomplished in 2 ways.

By the Fundamental Counting Principle (FCP) we can complete all 4 stages (and thus create a 3-person committee) in (4)(2)(2)(2) ways (= 32 ways)

Answer = E

Gmat_mission wrote: ↑

Wed Mar 24, 2021 7:17 am

A committee of three 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?

A. 16
B. 24
C. 26
D. 30
E. 32

Answer: E

Solution:

We are given that there are 4 married couples (or 8 people), and we need to determine the number of ways of choosing 3 people in which no 2 people are a married couple. So 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 is 8C3, which is calculated as follows:

8C3 = 8! / [3!(8-3)!] = (8 x 7 x 6) / 3! = 56

8, 7, and 6 in the numerator represent the number of ways in which 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 choice of 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.

Alternate Solution:

Without any restrictions, there are 8C3 = 8! / [3!(8-3)!] = (8 x 7 x 6) / 3! = 56 ways of choosing a committee of three people.

Let’s determine the number of committees that contain a couple. There are 4 choices for the couple. Once the couple is selected, the remaining spot can be filled by any one of the 6 remaining people. Thus, there are 4 x 6 = 24 committees that contain a married couple.

Since there are a total of 56 committees and since 24 of these committees contain a married couple, 56 - 24 = 32 committees do not contain a married couple.

Answer: E