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

Question

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.

bangalorian2000 · Answer

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

sookol · Answer

a small mistake 840 + 360 = 1200

careless mistakes are sometimes the most painful

nickk · Answer

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.

stne · Answer

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?

Bunuel · Answer

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.

TooLong150 · Answer

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?

oss198 · Answer

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?

oli29 · Answer

Why does this solution not work?

5C2*9C2*10C1

The last term represent the group which is left over and hence it doesn't matter if its an officer or a civilian?

bumpbot · Answer

Automated notice from GMAT Club BumpBot: 

A member just gave Kudos to this thread, showing it’s still useful. I’ve bumped it to the top so more people can benefit. Feel free to add your own questions or solutions.

 This post was generated automatically.

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

Prep Toolkit

Top 5 GMAT Debrief Videos

615 to 715 in 15 Days (Ria's Strategies)

More than Quant/Verbal Abilities: Speed vs. Accuracy

745 with Only Self-Prep and GMAT Club

From 555 to 765: 210-point Improvement

Perfect 805 Score Debrief by Julia

My Rewards

GMAT Problem Solving (PS) Questions

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

Prep Toolkit

615 to 715 in 15 Days (Ria's Strategies)

More than Quant/Verbal Abilities: Speed vs. Accuracy

745 with Only Self-Prep and GMAT Club

From 555 to 765: 210-point Improvement

Perfect 805 Score Debrief by Julia

My Rewards