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A five-member committee is to be formed from a group of five

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A five-member committee is to be formed from a group of five [#permalink] New post 11 Nov 2007, 14:37
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A five-member committee is to be formed from a group of five military officers and nine civilians. If the committee must include at least two officers and two civilians, in how many different ways can the committee be chosen?
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 [#permalink] New post 11 Nov 2007, 15:36
Incorrect, try again :)
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 [#permalink] New post 11 Nov 2007, 19:01
jimmyjamesdonkey wrote:
bingo, please explain reasoning.


Two possibilities:
1. 3 officers and 2 civilians: 5 C 3 * 9 C 2
2. 2 officers and 3 civilians: 5 C 2 * 9 C 3

Total possibilities = 5 C 3 * 9 C 2 + 5 C 2 * 9 C 3
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Re: PS: Perm & Combination Problem [#permalink] New post 27 Aug 2008, 17:46
x2suresh wrote:
jimmyjamesdonkey wrote:
A five-member committee is to be formed from a group of five military officers and nine civilians. If the committee must include at least two officers and two civilians, in how many different ways can the committee be chosen?


= 5 C 3 * 9 C 2 + 5 C 2 * 9 C 3
=1200


Question: The reason we donot divide the result by 14 C 5 as we are asked to find "how many different ways can the committee be chose" and not how many "different committees can be chosen?" Correct?
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Re: PS: Perm & Combination Problem [#permalink] New post 29 Mar 2009, 01:57
manOnFire wrote:
x2suresh wrote:
jimmyjamesdonkey wrote:
A five-member committee is to be formed from a group of five military officers and nine civilians. If the committee must include at least two officers and two civilians, in how many different ways can the committee be chosen?


= 5 C 3 * 9 C 2 + 5 C 2 * 9 C 3
=1200


Question: The reason we donot divide the result by 14 C 5 as we are asked to find "how many different ways can the committee be chose" and not how many "different committees can be chosen?" Correct?


No, I think if you ask "how many different committees can be chosen" you have to divide by the permutation of the chosen groups.

thus: Number of different committees:

Different committees of 3 officers / 2 civilians: (5 C 3 * 9 C 2) / 2! = (10 x 36) / 2 = 180

Different committes of 2 officers / 3 civilians: (5 C 2 * 9 C 3) / 2! = (10 x 84) / 2 = 420

Total committees: 180 + 420 = 600

this is analogous to: combination-55369.html

Walker, Suresh, is that correct?
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Re: PS: Perm & Combination Problem [#permalink] New post 29 Mar 2009, 02:59
What are the answer choices? It would be good to have those. And I am trying to understand why the answer is 600.
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Re: PS: Perm & Combination Problem [#permalink] New post 29 Mar 2009, 11:54
It's easy to read the question wrong (I think it's purposefully built this way), but once you realize there are 2 possibilities, the rest is easy:

2 officers and 3 civilians
or
3 officers and 2 civilians

9!/6!3!*5!/3!2! + 5!/2!3!*9!/7!2!=1200
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Re: PS: Perm & Combination Problem [#permalink] New post 17 Apr 2009, 19:53
peraspera wrote:
It's easy to read the question wrong (I think it's purposefully built this way), but once you realize there are 2 possibilities, the rest is easy:

2 officers and 3 civilians
or
3 officers and 2 civilians

9!/6!3!*5!/3!2! + 5!/2!3!*9!/7!2!=1200


that's exactly how i solved it. you gotta add the 2 diff possibilities
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Re: PS: Perm & Combination Problem [#permalink] New post 20 Apr 2009, 03:35
can anyone explain whats wrong with eyunni's reasoning
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Thanks
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Re: PS: Perm & Combination Problem [#permalink] New post 28 Sep 2009, 10:38
A five-member committee is to be formed from a group of five military officers and nine civilians. If the committee must include at least two officers and two civilians, in how many different ways can the committee be chosen?

Soln: 5C2 * 9C3 + 5C3 * 9C2
Re: PS: Perm & Combination Problem   [#permalink] 28 Sep 2009, 10:38
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