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A group consists of 5 officers and 9 civilians. If you must [#permalink]
24 Mar 2010, 08:11

00:00

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Difficulty:

(N/A)

Question Stats:

82% (02:08) correct
18% (02:44) wrong based on 7 sessions

A group consists of 5 officers and 9 civilians. If you must select a committee of 5 people, and the committee must include at least 2 officers and at least 2 civilians, how many different committees are possible?

Hi all,

A friend is taking a GMAT prep class and passed along this question to me - - sorry that I don't have the original source or the answer choices.

Re: tricky combinatorics problem [#permalink]
24 Mar 2010, 08:47

1

This post received KUDOS

gmatinfoonline wrote:

Hi all,

A friend is taking a GMAT prep class and passed along this question to me - - sorry that I don't have the original source or the answer choices.

"A group consists of 5 officers and 9 civilians. If you must select a committee of 5 people, and the committee must include at least 2 officers and at least 2 civilians, how many different committees are possible?"

Re: tricky combinatorics problem [#permalink]
26 Mar 2010, 03:51

I did it as follows (I'm pretty bad at combinatorics so please tell me where I went wrong):

5C2* 9C2 * 10C1 = 3600

We need at least 2 civilians and 2 officers. So first we choose 2 civilians from the 5, and then we choose 2 officers from the 9. We have one more place to fill, and 10 (7+3) people left, thus we choose 1 from 10.

Re: tricky combinatorics problem [#permalink]
09 Sep 2012, 04:21

bangalorian2000 wrote:

gmatinfoonline wrote:

Hi all,

A friend is taking a GMAT prep class and passed along this question to me - - sorry that I don't have the original source or the answer choices.

"A group consists of 5 officers and 9 civilians. If you must select a committee of 5 people, and the committee must include at least 2 officers and at least 2 civilians, how many different committees are possible?"

Re: tricky combinatorics problem [#permalink]
09 Sep 2012, 04:31

1

This post received KUDOS

Expert's post

stne wrote:

bangalorian2000 wrote:

gmatinfoonline wrote:

Hi all,

A friend is taking a GMAT prep class and passed along this question to me - - sorry that I don't have the original source or the answer choices.

"A group consists of 5 officers and 9 civilians. If you must select a committee of 5 people, and the committee must include at least 2 officers and at least 2 civilians, how many different committees are possible?"

please do confirm if my solution is right or wrong?

We are told that the committee must consist of 5 people, so the committee with 2 officers and 2 civilians is not possible.

A group consists of 5 officers and 9 civilians. If you must select a committee of 5 people, and the committee must include at least 2 officers and at least 2 civilians, how many different committees are possible?

To meet the conditions we can have only 2 cases: 2 officers and 3 civilians: \(C^2_5*C^3_9=840\); 3 officers and 2 civilians: \(C^3_5*C^2_9=360\);

Re: tricky combinatorics problem [#permalink]
09 Sep 2012, 04:54

Bunuel wrote:

stne wrote:

A friend is taking a GMAT prep class and passed along this question to me - - sorry that I don't have the original source or the answer choices.

"A group consists of 5 officers and 9 civilians. If you must select a committee of 5 people, and the committee must include at least 2 officers and at least 2 civilians, how many different committees are possible?"

please do confirm if my solution is right or wrong?

We are told that the committee must consist of 5 people, so the committee with 2 officers and 2 civilians is not possible.

A group consists of 5 officers and 9 civilians. If you must select a committee of 5 people, and the committee must include at least 2 officers and at least 2 civilians, how many different committees are possible?

To meet the conditions we can have only 2 cases: 2 officers and 3 civilians: \(C^2_5*C^3_9=840\); 3 officers and 2 civilians: \(C^3_5*C^2_9=360\);

Total: 840+360=1,200.

Hope it's clear.

yups that's clear , missed that logic completely , thank you _________________

Re: A group consists of 5 officers and 9 civilians. If you must [#permalink]
23 Oct 2013, 13:33

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Re: A group consists of 5 officers and 9 civilians. If you must [#permalink]
30 Oct 2013, 03:07

As Bunuel stated, there are two cases:

2 Officers, 3 Civilians (5 C 2) * (9 C 3) = 840

3 Officers, 2 Civilians (5 C 3) * (9 C 2) = 360

Total = 840+360 = 1200

Bunuel, it took me about 2 minutes to figure out the question, and 90 seconds to do the calculations. Are there any shortcuts to this type of question?

Re: tricky combinatorics problem [#permalink]
13 Nov 2013, 15:29

Bunuel wrote:

Hi all,

To meet the conditions we can have only 2 cases: 2 officers and 3 civilians: \(C^2_5*C^3_9=840\); 3 officers and 2 civilians: \(C^3_5*C^2_9=360\);

Total: 840+360=1,200.

Hope it's clear.

hi Bunuel, i read on this post six-highschool-boys-gather-at-the-gym-for-a-modified-game-161915.html that when the order does not matter, we divide by n!. Why in this case we don't divide \(C^2_5*C^3_9\) by 2!, as the order in which we choose the 2 officers and the 3 civilians doesn't matter?

gmatclubot

Re: tricky combinatorics problem
[#permalink]
13 Nov 2013, 15:29

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