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Director
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If a committee of 3 people is to be selected from among 5 [#permalink]
 06 Feb 2005, 11:23
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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
Enjoy 
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Manager
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# 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
D.
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Director
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Total number of ways - Number of ways that you end up selecing couples
10c3 - 5*1*8c1
= 80
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SVP
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D
Total ways to select committee - ways to select where we always have a couple
10C3 - 5x8C1 = 10x3x4 - 40 = 80
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Manager
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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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SVP
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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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GMAT Club Legend
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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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Manager
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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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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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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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Manager
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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
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