psm12se wrote:A committee of three people is to be chosen from four married couples. What is the number of different committee 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

Number of ways to choose 3 couples from 4 couples = 4C3 = 4

Now each of these 3 couples can send two persons (husband or wife), number of ways of doing this = 2 * 2 * 2 = 2^3 = 8

Therefore, total number of ways: 4C3 * 2^3 = 32

The correct answer is E.

psm12se wrote:A committee of three people is to be chosen from four married couples. What is the number of different committee 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

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, this stage can be accomplished in 4C3 ways (4 ways)

If anyone is interested, we have a free video on calculating combinations (like 4C3) in your head: https://www.gmatprepnow.com/module/gmat-counting?id=789

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

Cheers,
Brent

Aside: For more information about the FCP, we have a free video on the subject: https://www.gmatprepnow.com/module/gmat-counting?id=775

kevincanspain wrote:One more method to add to your repertoire:

Without the restriction concerning married couples, the number of committees consisting of 3 people would just be 8C3 = 8(7)(6)/3! = 56.
However, some of these have to be discarded, for they consist of an entire couple and a third person. How many must be discarded?

Good question.
We want to subtract all selections consisting of an entire couple and a third person. How many of selections are there?

To answer this question, let's break the task into stages.

Stage 1: Select 1 of the 4 couples.
We'll place both people in this couple on the committee.
There are 4 couples, so this stage can be accomplished in 4 ways

Stage 2: Select the third person for the committee
There are now 6 people remaining, so this stage can be accomplished in 6 ways.

By the Fundamental Counting Principle (FCP) we can complete both stages (and thus create a 3-person committee) in (4)(6) ways
In other words, we can create 24 committees that break the rule.

From here, when we take all 56 possible committees and subtract the 24 committees that break the rule, we get 32

Cheers,
Brent