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

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Senior Manager
Joined: 07 Nov 2004
Posts: 453
If a committee of 3 people is to be selected from among 5 [#permalink]

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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
Manager
Joined: 13 Oct 2004
Posts: 236

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06 Feb 2005, 14:14
# 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

D.
Director
Joined: 19 Nov 2004
Posts: 556
Location: SF Bay Area, USA

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06 Feb 2005, 17:01
Total number of ways - Number of ways that you end up selecing couples

10c3 - 5*1*8c1

= 80
VP
Joined: 18 Nov 2004
Posts: 1433

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06 Feb 2005, 17:07
D

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

10C3 - 5x8C1 = 10x3x4 - 40 = 80
Manager
Joined: 01 Jan 2005
Posts: 166
Location: NJ

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06 Feb 2005, 18:10

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.
VP
Joined: 18 Nov 2004
Posts: 1433

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06 Feb 2005, 18:59
Vijo wrote:

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.
GMAT Club Legend
Joined: 07 Jul 2004
Posts: 5043
Location: Singapore

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06 Feb 2005, 19:30
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
Manager
Joined: 01 Jan 2005
Posts: 166
Location: NJ

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06 Feb 2005, 20:55
banerjeea_98 wrote:
Vijo wrote:

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
VP
Joined: 18 Nov 2004
Posts: 1433

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06 Feb 2005, 21:32
Vijo wrote:
banerjeea_98 wrote:
Vijo wrote:

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.
SVP
Joined: 03 Jan 2005
Posts: 2233

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06 Feb 2005, 22:08
Vijo wrote:

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
Manager
Joined: 01 Jan 2005
Posts: 166
Location: NJ

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07 Feb 2005, 10:48
HongHu wrote:
Vijo wrote:

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
07 Feb 2005, 10:48
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