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A committee of three people is to be chosen from four [#permalink]

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27 Sep 2009, 23:41

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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?

There's two steps to this problem. Since each married couple can have a maximum of one representative, the first step is deciding which 3 of the 4 couples will be part of the committee. There are 4C3 ways to do this.

4C3 = 4

The second step is deciding which person in each married couple takes part in the committee. There are two people in each of the three married couples, so the final answer is:

For 1st person, there are 8 people can be chosen for 2nd person, there are 6 person can be chosen (1 have been chosen and his/her mate can not be chosen) for 3rd person, there are 4 person can be chosen This 3 people can be selected 3! different ways. So the solution is 8*6*4/3!=32.

Say three people selected w1, h2 and w4. These can be selected in these orders. w1h1w4 w1w4h1 h1w1w4 h1w4w1 w4w1h1 w4h1w1 Since we are dealing with combination not permutation. These 6 selections are same for us.

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

Each couple can send only one "representative" to the committee. Let's see in how many ways we can choose 3 couples (as there should be 3 members) out of 4 to send only one "representatives" to the committee: 4C3=4.

But each of these 3 couples can send two persons (husband or wife): 2*2*2=2^3=8.

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