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A five-member committee is to be formed from a group of five [#permalink]

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05 Mar 2005, 20:41

This topic is locked. If you want to discuss this question please re-post it in the respective forum.

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?

(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?

(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??????????

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

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.

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,