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Question

If a committee of 3 people is to be selected from among 5 married couples so that the committee does not include two people who are married to each other, how many such committees are possible?

EMPOWERgmatRichC · Answer

Hi Naveena1234,

There is a discussion of this question here:

https://gmatclub.com/forum/if-a-committ ... 88772.html

GMAT assassins aren't born, they're made,
Rich

Contact Rich at: [email protected]

Archit3110 · Answer

you can check the link shared in post above...
solution is
(10*8*6)/ 3! ; 80

abhikap14 · Answer

Lets consider the 5 couples as 5 entities. Now we have to select any 3 of these 5 entities. That will be 5C3
Now each of these selected entity has 2 options - Either the man of the woman can be in the committee. Hence 2 x 2 x 2.

Hence final answer is 5C3 x 2^3

ScottTargetTestPrep · Answer

We are given that there are five married couples (or 10 people), and we need to determine the number of ways of choosing 3 people in which no two 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.

If there are no restrictions, the number of ways of choosing 3 people from 10 people is 10C3, which is calculated as follows:

(10 x 9 x 8)/3! = 120

10, 9, and 8, 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 10 people. However, once a person is selected, that person’s spouse cannot be selected for the committee. This reduces to 8 the number of remaining people from which to select the second person (one person has already been selected, and that person’s spouse 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 6. Therefore, the number of ways of choosing these 3 people is:

(10 x 8 x 6)/3! = 80

Thus, there are 80 ways to choose such a committee.

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