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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 05 Mar 2005, 19:41
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?

a 119

b 1,200

c 3,240

d 3,600

e 14,400
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Re: comittee [#permalink] New post 05 Mar 2005, 19:48
=(5C2 x 9C3) + (5C3 x 9C2)=1200
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 [#permalink] New post 05 Mar 2005, 20:16
The OA is 1200.

The logic I used was:

(5C2)* (9C2) * (10C1) , but that equals 3600, which is not the correct answer.

5c2 (# of ways to pick 2 officers out of 5)
9c2 (# of ways to pick 2 civilians out of 9)
10c1 (# remaining people: 3officers+7civilians. Thus this should be the number of ways to pick one person out of the 10 remaining)

But if I compute above, I am getting the wrong answer. Can someone correct my logic?

Thanks
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 [#permalink] New post 05 Mar 2005, 20:25
[quote="cloaked_vessel"]The OA is 1200.


MA's answer is sufficient and very clean:
=(5C2 x 9C3) + (5C3 x 9C2)=

you either have 2 officers and 3 civilians or vice versa .. you don't need to calculate C from the "remaining" people
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Re: comittee [#permalink] New post 06 Mar 2005, 17:25
MA wrote:
=(5C2 x 9C3) + (5C3 x 9C2)=1200


:oops: i am so stupid, I've looking for the remaining one :lol:

14C5 - [(5C1*9C4) + (9C5) + (5C4*9C1) + (5C5)]

1 officer + 4 civilians
5 civilians
4 officers + 1 civilians
5 officers

As I am now trying to understand better the "comitee" problems and all about combinations, please tell me if the basic formula is correct or not..

thanks in advance
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 [#permalink] New post 11 Mar 2005, 14:21
I still don't understand why the answer isn't:

(5C2)* (9C2) * (10C1) = 3600

My reasoning is this:

From (5C2)* (9C2) you get: 360 groups

And you have 10 people left to pick so:

So here is the list of possibilities:

Group(1) -and- left-over-guy(1..10) gives 10 groups
Group(2) -and- left-over-guy(1..10) gives 10 groups
etc etc

so 360 * 10 = 3600
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 [#permalink] New post 11 Mar 2005, 17:26
cloaked_vessel wrote:
The OA is 1200.

The logic I used was:

(5C2)* (9C2) * (10C1) , but that equals 3600, which is not the correct answer.

5c2 (# of ways to pick 2 officers out of 5)
9c2 (# of ways to pick 2 civilians out of 9)
10c1 (# remaining people: 3officers+7civilians. Thus this should be the number of ways to pick one person out of the 10 remaining)

But if I compute above, I am getting the wrong answer. Can someone correct my logic?

Thanks


I used the same logic.. is this logic wrong??????????
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 [#permalink] New post 12 Mar 2005, 21:30
Please read this post, see if it helps. :)

http://www.gmatclub.com/phpbb/viewtopic ... 9520#89520
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 [#permalink] New post 12 Mar 2005, 21:40
Consider this: If I want to pick three person from four person, can I first pick two from the four, and then one from the remaining two? Let's see:

C(4,3)=4
C(4,2)*C(2,1)=6*2=12!!!

The difference betwee the two is the second approach would count
AB then C as different from AC then B, and different from BC then A. If you really want to get from approach one to two, you would have to change the whole thing to permultation, and then divide by the total permutation.

In other words you could do this:
P(4,2)*P(2,1)/P(3,3)=12*2/6=4
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 [#permalink] New post 12 Mar 2005, 21:45
We are not concerned with order, since we just need to get people in a team.

So, we have to choose:
- at least 2 offices to 2 civilians to form a 5 member team.

So the team could compris either (3 officers, 2 civilians), or (2 officers, 3 civilians)

group 1 (3 officers, 2 civilians)
# of combinations of 3 officers picked from 5 officers = 5!/3!2! = 10
# of combinations of 2 civilians picked from 9 civilians = 9!/2!7! = 36
So # of combinations of 3 officiers AND 2 civilians = 10*36 = 360

group 2 (2 officiers, 3 civilians)
# of combinations of 2 officers picked from 5 officers = 5!/3!2! = 10
# of combinations of 3 civilians picked from 9 civilians = 9!/3!6! = 84
So # of combinations of 2 officers AND 3 civilians = 10*84 = 840

Total number of combination = 360 +840= 1200 (b)

Hope this answers you queston. I've never solved probability questions with nCr or nPr or any other complicated hypergeometric distribution equations, etc.
In fact, the proability questions on the GMAT wouldn't require you to know them, you just need to know what you're looking for. But you are expected to be able to tell when order matters and when order does not matter, along with some basic counting rules. :-D
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 [#permalink] New post 13 Mar 2005, 11:36
ywilfred and HongHu thanks for your input!
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 [#permalink] New post 14 Jul 2005, 01:28
I think the question fails to mention atleast before the civilians it should read " at least two officers and atleast two civilians". Please note that there is no body else to choose from so there is no one person left. So it has to be,

5c3*9c2+ 5c2*9c3= 1200

THANKS :(
  [#permalink] 14 Jul 2005, 01:28
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