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Master | Next Rank: 500 Posts
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by aatech » Sat May 24, 2008 6:27 pm
A committee of three people is to be chosen from 4 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?

16, 24, 26, 30, 32
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by VP_Jim » Sat May 24, 2008 6:50 pm
This is a good one!

Rather than getting into formulas, let's think about it logically:

For the first person, we have 8 choices (we can pick anyone).

For the second person, we have 6 choices (we can pick anyone except the person already picked and his/her spouse).

For the third person, we have 4 choices (we can pick anyone except the two people already picked and their two spouses).

Thus, we have 192 different arrangements. But, since "order doesn't matter" in this case (e.g., a committee with Al, Bob, and Carl is the same as a committee with Carl, Bob, and Al), we have to divide.

Since we have three "items" in our calculation (the three committee members), we must divide 192 by 3! = 32.

Hope this helps!
Jim S. | GMAT Instructor | Veritas Prep
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by aatech » Sat May 24, 2008 6:52 pm
Understood.. OA is 32
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