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GMATPrep question

by rajesh_ctm » Thu Apr 26, 2007 8:05 pm
A committee of three people is to be chosen from four married couples. What is the number of different committees that can be chosen if two people who are married to each other cannot serve on the committee?

A 16
B 24
C 26
D 30
E 32
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by gm800 » Thu Apr 26, 2007 8:39 pm
Number of ways a 3 people committee can be formed from 8 people = 8c3

Number of committes in which a couple will serve together = 4 * 6c1 (i.e. 4 couples * 6 ways to select the 3rd person) = 24

So 8c3 - 24 = 32.
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by RAGS » Fri Apr 27, 2007 4:48 am
one couple is to be excluded
So selections of 3 from 6 people to be done

ie. 6C3 = 20
If It Is To Be It Is Up To Me
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by rajesh_ctm » Fri Apr 27, 2007 5:07 am
gm800, you got it right. Thanks for the explanation.

RAGS, you misinterpreted the question. Any two people married to each other can't be in the jury.
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by Cybermusings » Sat Apr 28, 2007 1:25 am
A committee of three people is to be chosen from four married couples. What is the number of different committees that can be chosen if two people who are married to each other cannot serve on the committee?

A 16
B 24
C 26
D 30
E 32

Number of ways in which 3 people can be chosen from a group of 8 people = 8C3 = 8*7*6/3*2*1 = 56
Number of ways in which 1 couple can be selected from a group of 4 = 4C1= 4
Number of ways in which 1 other person can be selected = 6C1 = 6
Total ways = 4*6 = 24
Now ways in which there are no couples = 56-24 = 32
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by f2001290 » Sat May 26, 2007 12:30 pm
Can someone clarify my approach?

In the above problem,

I can pick the 1st person in 8c1 ways.

I can pick the 2nd person in 6c1 ways. (I will not include the partner of 1st selection)

I can pick the 3rd person in 4c1 ways. (Eliminate partners of 1st and 2nd selections)

Total committes = 8c1*6c1*4c1

What is that I am missing here?
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by Cybermusings » Sat May 26, 2007 11:23 pm
A committee of three people is to be chosen from four married couples. What is the number of different committees that can be chosen if two people who are married to each other cannot serve on the committee?

A 16
B 24
C 26
D 30
E 32

Total committees possible (without restrictions) = 8C3 = 56 ways

Committees in which a couple does serve together = 4C1 (1 couple can be chosen from amongst 4 couples in 4C1 ways)...The third person can be chosen in 6C1 ways = 4C1*6C1 = 4*6 = 24

Hence committees in which couple do not serve together = 56-24 = 32
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