If a committee of 3 people is to be selected from among 5 ma : GMAT Problem Solving (PS)
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# If a committee of 3 people is to be selected from among 5 ma

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

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25 Jan 2008, 10:36
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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

OPEN DISCUSSION OF THIS QUESTION IS HERE: if-a-committee-of-3-people-is-to-be-selected-from-among-98533.html
[Reveal] Spoiler: OA

Last edited by Bunuel on 08 Dec 2013, 06:10, edited 1 time in total.
Renamed the topic, edited the question and added the OA.
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25 Jan 2008, 11:39
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(1) ->total combinations = 10c3 = 120
(2) ->total combinations which include atleast one couple = 5*8 = 40
(one couple can be choosen among the 5 couples in 5 ways..
the remaining one slot can be filled with any one of the 8 ppl remaining in 8 ways )

total combinations which do not include any couple = (1) - (2) = 80

another approach is
total combinations = (total permutations)/n!

= 10*8*6/3! = 480/6 = 80
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26 Jan 2008, 20:38
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netcaesar wrote:
If a committee of 3 people is to be selected from among 5 married couples so that the committee does not include 2 people who are married to each other, how many such committees are possible?

A) 20
B) 40
C) 50
D) 80
E) 120

First there are 10 people. and we want a committee of 3. So:

10!/3!7! --> 120 ways to have a comittee w/ no constraints.

120 - How many ways can we have couples?

lets say couple one is AA two BB three CC four DD and five EE
When we have a couple chosen, we have 8 others left so 8 different possible committees per couple

AAX (X stands for the other 8 members)
BBX
CCX
DDX
EEX

so 5*8=40 now just 120-40 = 80

D
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18 Aug 2008, 09:48
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

Can some explain how to solve this???
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18 Aug 2008, 09:51
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vivektripathi 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

D

Can some explain how to solve this???

Total number of ways to choose 3 people from 10: 10 choose 3 = 10!/(7!*3!) = 120
Total number of ways to include a married couple in the committee = 5*(8 choose 1) = 5*8 = 40

Total number of ways to not include a married couple = 120-40 = 80
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18 Aug 2008, 10:08
5C3 * (2^3) = 80

where
Total number of ways of choosing 3 out of 5 couples = 5C3 = 10
From each couple you can choose only one person. This can be done in 2 ways per couple. For 3 couples, 2^3 = 8 ways
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18 Aug 2008, 11:04
In the explanation
Total number of ways to choose 3 people from 10: 10 choose 3 = 10!/(7!*3!) = 120
Total number of ways to include a married couple in the committee = 5*(8 choose 1) = 5*8 = 40

Total number of ways to not include a married couple = 120-40 = 80

How do you get?

Total number of ways to include a married couple in the committee = 5*(8 choose 1) = 5*8 = 40
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18 Aug 2008, 11:41
rnemani wrote:
In the explanation
Total number of ways to choose 3 people from 10: 10 choose 3 = 10!/(7!*3!) = 120
Total number of ways to include a married couple in the committee = 5*(8 choose 1) = 5*8 = 40

Total number of ways to not include a married couple = 120-40 = 80

How do you get?

Total number of ways to include a married couple in the committee = 5*(8 choose 1) = 5*8 = 40

If you're going to form a committee with 1 couple and another person, then you have 5 couples to choose from.
There are 8 possibilities for the non-couple person from which you choose 1.
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18 Aug 2008, 12:43
cant this question be seen in this way (i know the answer comes wrong ) :

first select any person out of the 10 .

then we have 8 people to select the 3rd person from (excluding the pair above)

then for the 3rd person we have 6 choices (excluding the 2 pairs above)

ans = 10*8*6

(please lemme know why is this wrong ? )
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18 Aug 2008, 13:34
stallone wrote:
cant this question be seen in this way (i know the answer comes wrong ) :

first select any person out of the 10 .

then we have 8 people to select the 3rd person from (excluding the pair above)

then for the 3rd person we have 6 choices (excluding the 2 pairs above)

ans = 10*8*6

(please lemme know why is this wrong ? )

This isn't wrong, it's just incomplete.

You now have your Commitee of 3 people. But you've selected them in an order, when order doesn't matter.

Selection #1 = Jack
Selection #2 = Susan
Selection #3 = Mary

Assume this is a good committee where Jack is not married to Susan or Mary and Susan and Mary are not married. The above commitee will be the same as the following committee:

Selection #1 = Susan
Selection #2 = Mary
Selection #3 = Jack

To account for the fact that the order does not matter, we must divide out the number of ways a single committee can be the same members but in different order. 3 places = 3!, 3*2*1 = 6.

10*8*6 / 6 = 10*8 = 80. Answer D.
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13 Nov 2009, 14:55
I did answer the question correctly, however I am looking for a better approach to solve these kinds of question.
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13 Nov 2009, 18:57

4C3 ways to select the 3 couples.
2^3 ways to select the person from each couple (husband or wife)

4C3 * 2^3 = 4 * 8 = 32
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14 Nov 2009, 05:29
i would do it this way

Total possible combination = 8C3 = 56
Total Possible outcome which contain a couple : 4C1*6 (4 combination of couple plus one person from the rest) = 4*6 = 24

Ans = 56-24 = 32
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14 Nov 2009, 06:50
fireplayer

Can you elaborate your solution? Thanks
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26 Nov 2009, 04:11
fireplayer wrote:
i would do it this way

Total possible combination = 8C3 = 56
Total Possible outcome which contain a couple : 4C1*6 (4 combination of couple plus one person from the rest) = 4*6 = 24

Ans = 56-24 = 32

In all such problems we need to calculate the total possible combination. therefore as it was calculated above = 56.

We need to choose 3 people from 4 married couples, it will be the following scenarios.

a) 1 married couple and 1 single individual therefore if you selected one couple then 6 people are fighting for the last seat of the committee.

Possible outcome = 4C1 * 6 = 4*6 = 24

Number of possible committees = total - possible
= 56 - 24 = 32

More details you can find on the following url
http://www.beatthegmat.com/permutation-from-gmat-prep-t21950.html
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GMATPrep Married couples for a committee [#permalink]

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27 Nov 2009, 18:30
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28 Nov 2009, 06:27
There are 10C3 ways to select 3 people =>120 ways

Now there 5 ways to select a couple and the third person could be selected in 8 ways. So in total 40 ways

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08 Dec 2013, 06:01
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Re: If a committee of 3 people is to be selected from among 5 ma [#permalink]

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08 Dec 2013, 06:10
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Re: If a committee of 3 people is to be selected from among 5 ma [#permalink]

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08 Dec 2013, 06:11
netcaesar 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

Each couple can send only one "representative" to the committee. Let's see in how many ways we can choose 3 couples (as there should be 3 members) each to send one "representative" to the committee: 5C3=10.

But these 3 chosen couples can send two persons (either husband or wife): 2*2*2=2^3=8.

Total # of ways: 5C3*2^3=80.

OPEN DISCUSSION OF THIS QUESTION IS HERE: if-a-committee-of-3-people-is-to-be-selected-from-among-98533.html
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Re: If a committee of 3 people is to be selected from among 5 ma   [#permalink] 08 Dec 2013, 06:11
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