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Help with probability

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Help with probability

by liouedwin » Mon Jul 18, 2011 12:10 pm
Hey guys,

I'm stuck on a probability question and was hoping for some help. It would be great if someone can go over how to solve this particular problem!

A committee of three is chosen from four married couples. How many different committees can be formed if people who are married to each other cannot serve together on the committee?

The answer is 32, but I can't quite get it.

Thanks in advance!
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by gmatfeel » Mon Jul 18, 2011 1:09 pm
4 couples = 8 people in total.
The number of ways the committee can be formed out of 8 people =
= 8C3 (Using Combination formula)
= 8!/5!*3! = (Using Combination formula nCr = n!/(n-r)!*r!)
= 8*7 (solving and canceling the common terms from Numerator and denominator)
= 56

In choosing 3 people committee if one spot is taken by a couple, the third spot will be taken by the remaining people in 6C1 ways
So number of married couples we can have in each committee above is
= 4 * 6C1
= 4*6
= 24

Number of committees that can be formed without married couple serving together
= 56 - 24
= 32

That is a good question. Took lot of thinking.
Thanks...
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