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an easy question

by sana.noor » Wed Jul 31, 2013 8:55 am
If a committee of 4 people is to be selected from among 6 married couples so that the
committee includes four people who are married to each other, how many such
committees are possible?

isnt the answer 15? if we select teo couples from 6 as 6c2, we get the answer.
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by sana.noor » Wed Jul 31, 2013 9:00 am
here is another similar question

"If a committee of 4 people is to be selected from among 6 married couples so that the
committee includes exactly two people who are married to each other, how many such
committees are possible?"

now what should be the answer to this question, my answer is 60
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by GMATGuruNY » Wed Jul 31, 2013 10:38 am
sana.noor wrote:If a committee of 4 people is to be selected from among 6 married couples so that the
committee includes four people who are married to each other, how many such
committees are possible?

isnt the answer 15? if we select teo couples from 6 as 6c2, we get the answer.
A committee composed of 2 couples must include 2 men and 2 women.
From the 6 men, the number of ways to choose 2 = 6C2 = (6*5)/(2*1) = 15.
Number of options for the 2 women = 1. (The spouses of the 2 men must be selected.)
To combine these options, we multiply:
15*1 = 15.
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by GMATGuruNY » Wed Jul 31, 2013 11:09 am
sana.noor wrote:here is another similar question

"If a committee of 4 people is to be selected from among 6 married couples so that the
committee includes exactly two people who are married to each other, how many such
committees are possible?"

now what should be the answer to this question, my answer is 60
One married couple must be combined with one unmarried couple.

Married couple:
Number of options = 6.

Unmarried couple:
Once the married couple has been selected, 10 people remain.
From these 10 people, we must choose 2 who are unmarried.
Number of options for the first person selected = 10.
Number of options for the second person selected = 8. (Of the 9 remaining people, anyone but the first person's spouse.)
To combine these options, we multiply:
10*8.
Since the ORDER of the two people doesn't matter -- AB is same couple as BA -- we divide by the number of ways to ARRANGE the 2 people (2!):
(10*8)/(2*1) = 40.

To combine our options for each type of couple, we multiply the results above:
6*40 = 240.
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My students have been admitted to HBS, CBS, Tuck, Yale, Stern, Fuqua -- a long list of top programs.

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by vipulgoyal » Wed Jul 31, 2013 10:06 pm
for selecting 1 couple we have 6C1 options.
as we have selected 1 couple already, there are 5 couples left.
in these 5 couples we need to have 2 couples to select and 1 person from each couple.
so it will be 5C2 * 2 * 2.
ans = 6c1 * 5c2 * 2 * 2 = 240
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