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by Nidhs » Fri Feb 08, 2008 6:09 pm
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?


119

1,200

3,240

3,600

14,400

Can someone slove this for me?
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by luvaduva » Sat Feb 09, 2008 1:20 am
This is a combination problem, so you will be using the formula:

nCr = (n!)/(r!*(n-r)!)

You must break the problem up into two Cases. Find the combination of each part and then add them.

Case 1 is selecting exactly two officers and three civilians.

For the officers,

5C2 = (5!)/(2!*(5-2)!) = 10

For the civilians,

9C3 = (9!)/(3!*(9-3)!) = 84

So, 10*84 combinations for Case 1 = 840.

Case 2 is selecting exactly three officers and two civilians.

For the officers,

5C3 = (5!)/(3!*(5-3)!) = 10

For the civilians,

9C2 = (9!)/(2!*(9-2)!) = 36

So, 10*36 combinations for Case 2 = 360.

Adding Case 1 and Case 2 = 1200.

The answer is B.
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by GMATDUD » Sun Feb 10, 2008 11:55 am
I used a different approach and did get a different answer.

2 oficers from 5 can be selected in 5c2 ways = 10
2 civilians can be selected in 9c2 ways = 36
the remaining can be either officer or civilian and there are 10 people (7 civilians and 3 officers) left - that makes 10 c 1 ways = 10

hence 10*10*36 = 3600 ways.

Your approach seems logical Luv, but I am curious whats wrong in the approach I took .. should fit rite in with the requirement.

NIDS , whats the ans?
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by davidforsberg » Mon Feb 11, 2008 12:43 pm
I agree on this one with luvaduva, DUD, regarding your approach I don't think it's right because there are 7 Civilians and 3 Officers left, which would not give an equal probability between them, and from what I understand is that with C and V you cannot mix different variables. If I'm wrong, someone please correct me.
T minus 17 hours
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by GMATDUD » Mon Feb 11, 2008 8:52 pm
David,

we are only looking at # of ways in the third category and not looking at the probability. there are 10 folks left and we need 1 to fill the position . it could be either C or O. That was the reasoning in doing a 10C1.
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Answer is 3600

by senthil » Tue Feb 12, 2008 12:00 pm
Choosing the military officer as the fifth person coz there will be atleast two officers and two civilians in the five member committee.
5c2.9c2.3 ....( out of 5 officers two are selected rest is just 3 officers)
Choosing the civilians as the fifth person coz there will be atleast two officers and two civilians in the five member committee.
5c2.9c2.7 ....( out of 10civilians two are selected rest is just 7 civilians)

THEREFORE 5c2.9c2.3+5c2.9c2.7=5c2.9c2.10=3600ways

Thanks
Senthil
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by Stuart@KaplanGMAT » Tue Feb 12, 2008 2:28 pm
GMATDUD wrote:I used a different approach and did get a different answer.

2 oficers from 5 can be selected in 5c2 ways = 10
2 civilians can be selected in 9c2 ways = 36
the remaining can be either officer or civilian and there are 10 people (7 civilians and 3 officers) left - that makes 10 c 1 ways = 10

hence 10*10*36 = 3600 ways.

Your approach seems logical Luv, but I am curious whats wrong in the approach I took .. should fit rite in with the requirement.

NIDS , whats the ans?
The problem with this approach is that it will give a lot of duplicate solutions.

For example, you'll have combinations in which you chose officers A and B in your first selection and then officer C in your final selection. You'll have other combinations in which you chose officers A and C in your first selection and then officer B in your final selection. You've counted each of these as a separate possibility, when in fact they're identical.

Luvaduva's solution is perfect!
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