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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] New post 06 Feb 2005, 10:23
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A
B
C
D
E

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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

Enjoy :-D
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 [#permalink] New post 06 Feb 2005, 13:14
# of ways to pick 3 from 10 = 10C3 = 120
# of ways to pick couples in 3 = 5.8 = 40
# of ways that no couples in 3 = 120 - 40 = 80

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 [#permalink] New post 06 Feb 2005, 16:01
Total number of ways - Number of ways that you end up selecing couples

10c3 - 5*1*8c1

= 80
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 [#permalink] New post 06 Feb 2005, 16:07
D

Total ways to select committee - ways to select where we always have a couple

10C3 - 5x8C1 = 10x3x4 - 40 = 80
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 [#permalink] New post 06 Feb 2005, 17:10
I thought about this question in an entirely different manner

For 3 ppl,

I have 10 choices to pick 1st person

I have 8 choices to pick 2nd person(excluding 1st person n his/her spouse)
I have 6 choices to pick 3rd person (excluding 1st person n his/her spouse and 2nd person n his spouse)


Thus 10*8*6=480.

Pls explain.
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 [#permalink] New post 06 Feb 2005, 17:59
Vijo wrote:
I thought about this question in an entirely different manner

For 3 ppl,

I have 10 choices to pick 1st person

I have 8 choices to pick 2nd person(excluding 1st person n his/her spouse)
I have 6 choices to pick 3rd person (excluding 1st person n his/her spouse and 2nd person n his spouse)


Thus 10*8*6=480.

Pls explain.


Order of picking doesn't matter, so have to use combination instead of permutation.
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 [#permalink] New post 06 Feb 2005, 18:30
possibilites of 3 people team = 10!/3!7! = 120 teams

possbilites that inside 3 people team, 2 are married to each other
= 3!/2! = 3

teams consisting of 1 married couple = 8*5 = 40

teams consisting of no married couples = 120-40=80
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 [#permalink] New post 06 Feb 2005, 19:55
banerjeea_98 wrote:
Vijo wrote:
I thought about this question in an entirely different manner

For 3 ppl,

I have 10 choices to pick 1st person

I have 8 choices to pick 2nd person(excluding 1st person n his/her spouse)
I have 6 choices to pick 3rd person (excluding 1st person n his/her spouse and 2nd person n his spouse)


Thus 10*8*6=480.

Pls explain.


Order of picking doesn't matter, so have to use combination instead of permutation.


Banerjee I used combinations : 10C1 * 8C1 * 6C1 = 10 * 8 * 6
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 [#permalink] New post 06 Feb 2005, 20:32
Vijo wrote:
banerjeea_98 wrote:
Vijo wrote:
I thought about this question in an entirely different manner

For 3 ppl,

I have 10 choices to pick 1st person

I have 8 choices to pick 2nd person(excluding 1st person n his/her spouse)
I have 6 choices to pick 3rd person (excluding 1st person n his/her spouse and 2nd person n his spouse)


Thus 10*8*6=480.

Pls explain.


Order of picking doesn't matter, so have to use combination instead of permutation.


Banerjee I used combinations : 10C1 * 8C1 * 6C1 = 10 * 8 * 6


10C1 = 10P1 = 10......u have to create select group and not arrange ppl in this ques.
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 [#permalink] New post 06 Feb 2005, 21:08
Vijo wrote:
I thought about this question in an entirely different manner

For 3 ppl,

I have 10 choices to pick 1st person

I have 8 choices to pick 2nd person(excluding 1st person n his/her spouse)
I have 6 choices to pick 3rd person (excluding 1st person n his/her spouse and 2nd person n his spouse)


Thus 10*8*6=480.

Pls explain.


You are in fact doing permulation. In your method, if you pick A first, and then B second, and then C third, this would count as a different choice from if you pick B first, then A, then C, also different from C, A, B ... etc.

Therefore you need one more step from your way of thinking. You know that you have got the number of permulations, and you need to get the number of combinations. You just need to divide your number by the total permulation of the three chosen people, since P(m,n)=C(m,n)*N!

Therefore you get 480/3!=80
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 [#permalink] New post 07 Feb 2005, 09:48
HongHu wrote:
Vijo wrote:
I thought about this question in an entirely different manner

For 3 ppl,

I have 10 choices to pick 1st person

I have 8 choices to pick 2nd person(excluding 1st person n his/her spouse)
I have 6 choices to pick 3rd person (excluding 1st person n his/her spouse and 2nd person n his spouse)


Thus 10*8*6=480.

Pls explain.


You are in fact doing permulation. In your method, if you pick A first, and then B second, and then C third, this would count as a different choice from if you pick B first, then A, then C, also different from C, A, B ... etc.

Therefore you need one more step from your way of thinking. You know that you have got the number of permulations, and you need to get the number of combinations. You just need to divide your number by the total permulation of the three chosen people, since P(m,n)=C(m,n)*N!

Therefore you get 480/3!=80




Thanks a bunch Honghu. I got it....where was I going wrong.

Vijo
  [#permalink] 07 Feb 2005, 09:48
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