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committee

by Mayur Sand » Sat Jul 25, 2009 5:09 am
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?

A. 20
B. 40
C. 50
D. 80
E. 120

oa D
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by shibal » Sat Jul 25, 2009 10:26 am
(10*8*6)/3!
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by Mayur Sand » Sat Jul 25, 2009 10:59 am
You mind explaining how you got that equation
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Re: committee

by Stuart@KaplanGMAT » Sat Jul 25, 2009 11:14 am
Mayur Sand wrote: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?

A. 20
B. 40
C. 50
D. 80
E. 120
For our first person, we have 10 choices.

Once we choose that person, we also must eliminate his or her spouse. So, for our second person, we have 8 choices.

Once we choose that person, we also must eliminate his or her spouse. So, for our third person, we have 6 choices.

So far we have 10*8*6.

However, we need to recognize that using this method of calculation we will have counted the same committees multiple times.

For example, we could have chosen Bob, then Fred, then Sam. We could also have chosen Fred, then Bob, then Sam. Even though both options give us the exact same committee, we've counted that committee twice.

So, we need to factor out the duplicates. There are 3! ways to arrange 3 people, so that's what we need to factor out of our calculation.

Accordingly, our final answer is:

10*8*6/3! = 10*8*6/6 = 10*8 = 80... choose (D).
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