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

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If a committee of 3 people is to be selected from among 5 [#permalink]  26 Dec 2005, 08:53
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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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Beat the tough questions...

VP
Joined: 22 Aug 2005
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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
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Re: DS- Committee 3 People - Tough one [#permalink]  26 Dec 2005, 12:17
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
_________________

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

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Re: DS- Committee 3 People - Tough one [#permalink]  26 Dec 2005, 16:40
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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Explain [#permalink]  29 Dec 2005, 22:17
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
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Posts: 1059
Location: CA
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Re: Explain [#permalink]  29 Dec 2005, 22:21
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, 22:21
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 It is currently 06 Jul 2015, 23:28

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

Author Message
TAGS:
Manager
Joined: 10 Oct 2005
Posts: 113
Location: Hollywood
Followers: 0

Kudos [?]: 16 [0], given: 0

If a committee of 3 people is to be selected from among 5 [#permalink]  26 Dec 2005, 08:53
00:00

Difficulty:

(N/A)

Question Stats:

0% (00:00) correct 0% (00:00) wrong based on 0 sessions
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
_________________

The GMAT, too tough to be denied.
Beat the tough questions...

VP
Joined: 22 Aug 2005
Posts: 1120
Location: CA
Followers: 1

Kudos [?]: 49 [0], given: 0

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
_________________

Whether you think you can or think you can't. You're right! - Henry Ford (1863 - 1947)

VP
Joined: 06 Jun 2004
Posts: 1059
Location: CA
Followers: 2

Kudos [?]: 54 [0], given: 0

Re: DS- Committee 3 People - Tough one [#permalink]  26 Dec 2005, 12:17
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
_________________

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

SVP
Joined: 05 Apr 2005
Posts: 1731
Followers: 5

Kudos [?]: 39 [0], given: 0

Re: DS- Committee 3 People - Tough one [#permalink]  26 Dec 2005, 16:40
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
Followers: 0

Kudos [?]: 0 [0], given: 0

Explain [#permalink]  29 Dec 2005, 22:17
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: 1059
Location: CA
Followers: 2

Kudos [?]: 54 [0], given: 0

Re: Explain [#permalink]  29 Dec 2005, 22:21
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, 22:21
Similar topics Replies Last post
Similar
Topics:
11 From a group of 10 people, a committee of four people is to 5 20 Nov 2013, 02:12
2 Sets A and B each consist of three terms selected from the 5 3 25 Dec 2012, 23:33
8 Company X is forming a two-person committee by selecting one 18 20 Sep 2011, 20:12
26 A committee of 2 people is to be formed from a group of 8 11 22 Jan 2011, 02:06
A set {1,2,3,4,5,6}. select 2 numbers from the set, is the 2 17 Mar 2006, 08:33
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