Yurik79 wrote:

Professor wrote:

ffgmat wrote:

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?

1. 16

2. 24

3. 26

4. 30

5. 32

= 8c3 - 4x6

= 32

Prof please explain what is 4x6.I am stuck with this part

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3 member committee can be formed

only in two ways:

1. All three are unrelated. Say X

2. 1 couple + another person. Say Y

Total number of ways to form 3 member committee can be formed = 8c3

Hence 8c3 = X + Y

Y = (select one couple) * (select one person from remaining 6) = 4.6

So, X = 8c3 - 4.6

Hope this clarifies