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Married couples

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Married couples

by crackgmat007 » Fri May 22, 2009 8:04 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
Last edited by crackgmat007 on Thu May 28, 2009 9:59 am, edited 1 time in total.
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by bkeathley » Fri May 22, 2009 9:20 am
You have 10 choices for the first committee spot. You cannot select the first choice, or their spouse for the second so you have 8 choices for the second committee spot. You still cannot select the first selection or their spouse, and now you cannot select the second selection or their spouse so you have 6 choices for the third committee spot.

That is 10 * 8 * 6 choices. Order doesn't matter on this committee because there is no difference between the 3 committee spots so we divide out by the 3! that accounts for the ordering and get 10*8*6/3! = 10*8*6/6 = 10 * 8 = 80. The answer is D.
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Re: Married couples

by dtweah » Fri May 22, 2009 9:39 am
crackgmat007 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

OA - D
Split the couples into two groups of unmarried couples. This changes the problem to how many ways of choosing 3 people from two groups of 5 each with the given restrictions

ABCDE FGHIJ

AF are married BG are married.....


5C3 + 5C3= 20 (3 from either group)
5C2 x 3C1 =30 (2 from G1 and 1 from G2: If you are choosing 2 persons from 1 that eliminates 2 persons automatically from Group 2 so you can only choose 1 person from 3)
5C1 x 4C2= 30 ( 1 from group 1 eliminates 1 from 2 leaving 2 choices from 4)
The order of choice does not mater so this should give 80.

Choose D.
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is this correct

by rah_pandey » Thu May 28, 2009 4:20 am
we have 5 couples. choose any 3. now this can be done in 5c3 ways. Now we have to pick one person from these 3 chosen couple groups. 2c1*2c1*2c1=>10*8=80
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by ghacker » Wed Jun 10, 2009 10:20 am
Suppose there are three spaces A B C we can , so there are 6 (3!) ways of arranging the 3 places

A can be filled in 10 ways , B can be filled in 8 ways and C can be filled in 6 ways

So the total number of ways = 10*8*6/3! = 80
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