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# GMATprep practice question: can somebody please solve it

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Manager
Joined: 05 Jan 2009
Posts: 81

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25 Apr 2009, 15:43
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Could you please solve it with explanation in each step?

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

[Reveal] Spoiler:
OA is 32
Intern
Joined: 18 Feb 2009
Posts: 5

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25 Apr 2009, 22:33
I got E.

since no married couple is allowed, consider 4 couples as 4 element. So, we need to select the 3 couples from 4 and then who among the selected couple will be part of the group.

3 couple can be selected in 4c3 ways and a member can be selected in 2 ways from each of the selected couple.

So, total number of ways will be = 4c3*2*2*2 = 32

Thanks
Akshay
Manager
Joined: 05 Jan 2009
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26 Apr 2009, 14:53
Why are you doing 2*2*2? can you please explain?
Manager
Joined: 22 Jul 2009
Posts: 191

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21 Sep 2009, 13:37
. Four married couples = 8 people
. 3 spots

8 * 6 * 4 = number of possible permutations

But here order does not matter => we want the number of possible combinations, not permutations.

nPk = nCk * k!

=> 8 * 6 * 4 / 3! = 32
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Last edited by powerka on 22 Sep 2009, 06:16, edited 1 time in total.
Manager
Joined: 22 Jul 2009
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21 Sep 2009, 13:40
pmal04 wrote:
Why are you doing 2*2*2? can you please explain?

4C3 * 2! * 2! * 2! is an alternative approach, equally valid.

That's picking 3 people out of 4 couples, and then accounting for the possible combinations within each couple (each couple has two people, so possible combinations within each couple are 2!).
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Intern
Joined: 02 Sep 2009
Posts: 8

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21 Sep 2009, 14:28
I got E as well.

My answer may not be as efficient as some but here is what I did.

Step 1: I started with 8 Choose 3 - since we have a total of 8 and are picking a group of 3. Choose Combination instead of Permutations since order does not matter. 8 Choose 3 = 56

Step 2: I know that x<56 since some of the combinations above will yield 2 people from the same married group. So next I figured out how many combinations would yield a married couple in the group. I thought of the population as follows to do so quickly;

A1 A2 B1 B2 C1 C2 D1 D2 -

If there is a married couple in 1 group and only 1 spot left I deduced that it could only be 6 different people in with each married couple. So 4 married couples x 6 available options yields 24 groups which have a married couple (6x4=24)

Step 3: 56 - 24 = 32
Re: GMATprep practice question: can somebody please solve it   [#permalink] 21 Sep 2009, 14:28
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