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If a committee of 3 people is to be selected from among 5

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Manager
Joined: 10 Oct 2005
Posts: 113
Location: Hollywood
If a committee of 3 people is to be selected from among 5 [#permalink]

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26 Dec 2005, 09:53
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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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VP
Joined: 22 Aug 2005
Posts: 1112
Location: CA

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26 Dec 2005, 10:01
D. 80

1st member can be selected out of 10 people = 10
2nd member can be selected from 10 - 2 = 8 (remove first member and his/her spouse) = 8
3rd member : 10 - 4 = 6

total: 10 * 8 * 6
remove duplicates as order does not matter:
10 *8 * 6 / 3! = 80
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VP
Joined: 06 Jun 2004
Posts: 1053
Location: CA
Re: DS- Committee 3 People - Tough one [#permalink]

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26 Dec 2005, 13:17
TOUGH GUY 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

Total comb. for a committee of 3 - Total comb. a committee of 3 including a married couple

10C3 - (5C1*8C1)

120 - (5*8) = 80
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SVP
Joined: 05 Apr 2005
Posts: 1710
Re: DS- Committee 3 People - Tough one [#permalink]

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26 Dec 2005, 17:40
TeHCM wrote:
Total comb. for a committee of 3 - Total comb. a committee of 3 including a married couple
=10C3 - (5C1*8C1)
=120 - (5*8) = 80

this is also my approach.......
Intern
Joined: 28 Dec 2005
Posts: 3

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29 Dec 2005, 23:17
Total comb. for a committee of 3 - Total comb. a committee of 3 including a married couple

10C3 - (5C1*8C1)

Please elaborate 5C1*8C1 in the above explaination?
VP
Joined: 06 Jun 2004
Posts: 1053
Location: CA

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29 Dec 2005, 23:21
Yogendras wrote:
Total comb. for a committee of 3 - Total comb. a committee of 3 including a married couple

10C3 - (5C1*8C1)

Please elaborate 5C1*8C1 in the above explaination?

There are 5 couples and I want to know how many ways I could pick out one couple....thus 5C1.

Since two couples are picked, I want to know how many ways I could pick the 1 out of the people that's left...thus (10-2)C1
_________________

Don't be afraid to take a flying leap of faith.. If you risk nothing, than you gain nothing...

Re: Explain   [#permalink] 29 Dec 2005, 23:21
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