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

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Senior Manager
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If a committee of 3 people is to be selected from among 5 [#permalink]

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05 May 2006, 20:42
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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
VP
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Re: PS: committees of married couples. [#permalink]

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05 May 2006, 21:52
M8 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 = 10c3 = 120
no of ways 3 people with 2 couple and a single can be selected = 5 x8 = 40
so, no of ways 3 people with all individual singles can be selected = 120 - 40 = 80
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05 May 2006, 22:32
Agree with 80
VP
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Re: PS: committees of married couples. [#permalink]

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06 May 2006, 13:15
M8 wrote:
Professor wrote:
total = 10c3 = 120
no of ways 3 people with 2 couple and a single can be selected = 5 x8 = 40
so, no of ways 3 people with all individual singles can be selected = 120 - 40 = 80

Prof what is 5*8?

1. take one couple at a time and then also take one from the rest 8 people = 1 x 8 = 8 group
2. second couple = 8
.
.
.
.
5 fifth couple = 8

total = 40
so the desired combination = total - no of comb with a couple = 120-40=80
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10 May 2006, 03:56
Thanks Prof, the OA is 'D' - 80.
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10 May 2006, 06:25
There are 10 people to choose from for the first position, then 8 for the second (the 10 that start less the one chosen and their spouse), then 6 for the third choice. And these 3 people can be rearranged 3! ways.

# = 10*8*6/3!
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10 May 2006, 06:40
mattflow wrote:
There are 10 people to choose from for the first position, then 8 for the second (the 10 that start less the one chosen and their spouse), then 6 for the third choice. And these 3 people can be rearranged 3! ways.

# = 10*8*6/3!

Exactly. Just to clarify the last point, we have to divide by 3! because
the order of the people in the committee doesn't matter.
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Re: PS: committees of married couples. [#permalink]

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10 May 2006, 09:48
Professor wrote:
M8 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 = 10c3 = 120
no of ways 3 people with 2 couple and a single can be selected = 5 x8 = 40
so, no of ways 3 people with all individual singles can be selected = 120 - 40 = 80

Nice explanation prof
However, I was wondering..what is wrond wid....10C1 * 8C1 * 6C1
10C1---> take one person out of 10
8C1-->leave the spouse of the person selected in above round..
6C1--->leave the spuse of pwerson selected above...

Plz tell what m I missing..
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11 May 2006, 23:09
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Agree with D 80
For the 1 person there are 10 choices, for the 2 – 8, for the 3rd – 6. 10x8x6=480. Since, the order doesn’t matter (abc=cba), divide it by 3! And get 80.
11 May 2006, 23:09
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# If a committee of 3 people is to be selected from among 5

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