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Senior | Next Rank: 100 Posts
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by duke » Sun Jul 27, 2008 12:09 am
Pls help me understand this Q correctly. THx.

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
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by preetha_85 » Sun Jul 27, 2008 12:41 am
Hi

We need 3 people from the 5 couples.
So we will first select 3 couples out of the 5 i.e 5C3 ways.
Now from each of the couple selected we can only select one of them this can be done in 2C1 ways.

So the answer is : 5C3*2C1*2C1*2C1 = 80.
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by pepeprepa » Sun Jul 27, 2008 12:46 am
Here is how I see the problem

1)
The probability not to have two married in the group is 2/3
( P=1*( 8/9 )*( 6/8 )=6/9 )

2)
The total number of possibilities is 120

3)
Cross multiply 2/3=x/120
x=80
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by duke » Sun Jul 27, 2008 12:49 am
thanks all. 80 is correct one.
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