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by eaakbari » Sat Nov 10, 2012 8:39 am
I have solved this problem using logic and got the answer as 32, but can someone give the approach to this question using the formula for Combinations.

A committee of 3 people is to be selected from a group of 8 people, which includes 4 married couples. If the committee cannot contain more than one member of any married couple, how many 3-person committees are possible?
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by GMATGuruNY » Sat Nov 10, 2012 8:54 am
A committee of three people is to be chosen from four married couples. What is the number of different commities that can be chosen if two people who are married to each other cannot both serve on this committee?

16
24
26
30
32

OA: E
One approach:

Number of options for the 1st person = 8.
Number of options for the 2nd person = 6. (Of the 7 people left, we can't use the mate of the 1st person chosen, leaving 7-1= 6 choices.)
Number of options for the 3rd person = 4. (Of the 6 people left, we can't use the mates of the 2 people already chosen, leaving 6-2 = 4 choices.)

Since the ORDER of the selections doesn't matter -- ABC is the same COMMITTEE as BCA -- we divide by the number of ways to arrange the 3 people chosen:
(8*6*4)/3! = 32.

The correct answer is E.
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