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# A group consists of 5 officers and 9 civilians. If you must

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A group consists of 5 officers and 9 civilians. If you must [#permalink]  24 Mar 2010, 08:11
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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.
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Re: tricky combinatorics problem [#permalink]  24 Mar 2010, 08:47
1
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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?"

2 officers & 3 civilans = 5C2 * 9C3
3 officers & 2 civilans = 5C3 * 9C2
total = 5C2 * 9C3 + 5C3 * 9C2 = 840 + 360 = 1200

Last edited by bangalorian2000 on 26 Mar 2010, 05:54, edited 1 time in total.
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Re: tricky combinatorics problem [#permalink]  25 Mar 2010, 16:18
bangalorian2000 wrote:
total = 5C2 * 9C3 + 5C3 * 9C2 = 840 + 360 = 1170

a small mistake 840 + 360 = 1200

careless mistakes are sometimes the most painful
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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.
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Re: tricky combinatorics problem [#permalink]  26 Mar 2010, 05:54
sookol wrote:
bangalorian2000 wrote:
total = 5C2 * 9C3 + 5C3 * 9C2 = 840 + 360 = 1170

a small mistake 840 + 360 = 1200
careless mistakes are sometimes the most painful

Thanks sookol for pointing mistake.
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Re: tricky combinatorics problem [#permalink]  29 Mar 2010, 12:15
Thanks for the explanation! The correct answer is indeed 1200.
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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?"

2 officers & 3 civilans = 5C2 * 9C3
3 officers & 2 civilans = 5C3 * 9C2
total = 5C2 * 9C3 + 5C3 * 9C2 = 840 + 360 = 1200

shouldn't there one more case here, the case when exactly 2 officers and 2 civilians are selected
$$C^5_2 *C^9_2 * C^{10}_1$$ = 3600

so total ways

2 officers 2 civilians = $$C^5_2 *C^9_2 * C^{10}_1$$ = 3600
3 officers 2 civilians = $$C^5_3 * C^9_2$$ = 840
2 officers 3 civilians = $$C^5_2 * C^9_3$$= 360

so 3600 +1200 = 4800

please do confirm if my solution is right or wrong?
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Re: tricky combinatorics problem [#permalink]  09 Sep 2012, 04:31
1
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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?"

2 officers & 3 civilans = 5C2 * 9C3
3 officers & 2 civilans = 5C3 * 9C2
total = 5C2 * 9C3 + 5C3 * 9C2 = 840 + 360 = 1200

shouldn't there one more case here, the case when exactly 2 officers and 2 civilians are selected
$$C^5_2 *C^9_2 * C^{10}_1$$ = 3600

so total ways

2 officers 2 civilians = $$C^5_2 *C^9_2 * C^{10}_1$$ = 3600
3 officers 2 civilians = $$C^5_3 * C^9_2$$ = 840
2 officers 3 civilians = $$C^5_2 * C^9_3$$= 360

so 3600 +1200 = 4800

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.
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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?"

2 officers & 3 civilans = 5C2 * 9C3
3 officers & 2 civilans = 5C3 * 9C2
total = 5C2 * 9C3 + 5C3 * 9C2 = 840 + 360 = 1200

shouldn't there one more case here, the case when exactly 2 officers and 2 civilians are selected
$$C^5_2 *C^9_2 * C^{10}_1$$ = 3600

so total ways

2 officers 2 civilians = $$C^5_2 *C^9_2 * C^{10}_1$$ = 3600
3 officers 2 civilians = $$C^5_3 * C^9_2$$ = 840
2 officers 3 civilians = $$C^5_2 * C^9_3$$= 360

so 3600 +1200 = 4800

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