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# A committee of three people is to be chosen from four

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Manager
Joined: 13 Mar 2009
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A committee of three people is to be chosen from four [#permalink]

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28 Sep 2009, 00:41
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64% (01:47) correct 36% (01:25) wrong based on 119 sessions

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A committee of three people is to be chosen from four married couples. What is the number of different committees that can be chosen if two people who are married to each other cannot both serve on the committee?

A. 16
B. 24
C. 26
D. 30
E. 32

OPEN DISCUSSION OF THIS QUESTION IS HERE: a-committee-of-3-people-is-to-be-chosen-from-four-married-94068.html
[Reveal] Spoiler: OA

Last edited by Bunuel on 06 Aug 2012, 11:18, edited 3 times in total.
Edited the question and added the OA.
Manager
Joined: 11 Sep 2009
Posts: 129
Re: Combinatorics form GMATPrep [#permalink]

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28 Sep 2009, 00:46
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The answer is E: 32.

There's two steps to this problem. Since each married couple can have a maximum of one representative, the first step is deciding which 3 of the 4 couples will be part of the committee. There are 4C3 ways to do this.

4C3 = 4

The second step is deciding which person in each married couple takes part in the committee. There are two people in each of the three married couples, so the final answer is:

4C3 * (2 * 2 * 2) = 32
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Joined: 23 Jun 2009
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Location: Turkey
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Re: Combinatorics form GMATPrep [#permalink]

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28 Sep 2009, 00:49
For 1st person,
there are 8 people can be chosen
for 2nd person,
there are 6 person can be chosen (1 have been chosen and his/her mate can not be chosen)
for 3rd person,
there are 4 person can be chosen
This 3 people can be selected 3! different ways.
So the solution is
8*6*4/3!=32.
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Joined: 23 Jun 2009
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Re: Combinatorics from GMATPrep [#permalink]

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28 Sep 2009, 00:50
Another way.
We can select 3 couples in 4 couples. And select each mate with 2 different ways. So the solution is
4C3*2*2*2=32
Senior Manager
Joined: 31 Aug 2009
Posts: 417
Location: Sydney, Australia
Re: Combinatorics form GMATPrep [#permalink]

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28 Sep 2009, 06:35
maliyeci wrote:
For 1st person,
This 3 people can be selected 3! different ways.
So the solution is
8*6*4/3!=32.

Hi Maliyeci... can you explain this part for me?
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Re: Combinatorics from GMATPrep [#permalink]

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28 Sep 2009, 06:46
Exactly.

Say three people selected w1, h2 and w4. These can be selected in these orders.
w1h1w4
w1w4h1
h1w1w4
h1w4w1
w4w1h1
w4h1w1
Since we are dealing with combination not permutation. These 6 selections are same for us.
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Re: Combinatorics from GMATPrep [#permalink]

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28 Sep 2009, 06:55
nice explanation. Thanks
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Re: Combinatorics from GMATPrep [#permalink]

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02 May 2011, 23:16
4c3 * (2*2*2) = 32 is also a good way to think.
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Re: Combinatorics from GMATPrep [#permalink]

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06 Aug 2012, 11:10
total-8 ppl
8*6*4/3!=32
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Posts: 39720
Re: A committee of three people is to be chosen from four [#permalink]

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06 Aug 2012, 11:19
A committee of three people is to be chosen from four married couples. What is the number of different committees that can be chosen if two people who are married to each other cannot both serve on the committee?

A. 16
B. 24
C. 26
D. 30
E. 32

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) out of 4 to send only one "representatives" to the committee: 4C3=4.

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

Total # of ways: 4C3*2^3=32.

OPEN DISCUSSION OF THIS QUESTION IS HERE: a-committee-of-3-people-is-to-be-chosen-from-four-married-94068.html
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Re: A committee of three people is to be chosen from four   [#permalink] 06 Aug 2012, 11:19
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# A committee of three people is to be chosen from four

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