Last visit was: 28 May 2024, 18:29 It is currently 28 May 2024, 18:29
Decision Tracker
My Rewards
Unanswered
Active Topics
New posts
New comers' posts
Close
GMAT Club Daily Prep
Thank you for using the timer - this advanced tool can estimate your performance and suggest more practice questions. We have subscribed you to Daily Prep Questions via email.

Customized
for You

we will pick new questions that match your level based on your Timer History

Track
Your Progress

every week, we’ll send you an estimated GMAT score based on your performance

Practice
Pays

we will pick new questions that match your level based on your Timer History
Not interested in getting valuable practice questions and articles delivered to your email? No problem, unsubscribe here.
Go to My Error Log Learn more
Close
Request Expert Reply
Confirm Cancel

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

SORT BY:
Date
Kudos
Tags:
Find Similar Topics 
Show Tags
Hide Tags
avatar
gmatinfoonline
Intern
Intern
Joined: 19 Mar 2010
Posts: 2
Own Kudos [?]: 11 [7]
Given Kudos: 3
Location: San Diego, CA
Concentration: MBA, Finance, 2009
Schools:UCLA Anderson
Send PM
A group consists of 5 officers and 9 civilians. If you must [#permalink] A group consists of 5 officers and 9 civilians. If you must   24 Mar 2010, 09:11
7
Bookmarks
00:00
A
B
C
D
E

Difficulty:

(N/A)

Question Stats:

89% (01:44) correct 11% (02:44) wrong based on 48 sessions
Hide Show timer Statistics
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.
User avatar
bangalorian2000
Manager
Manager
Joined: 01 Feb 2010
Posts: 90
Own Kudos [?]: 136 [1]
Given Kudos: 2
Send PM
Re: tricky combinatorics problem [#permalink] A group consists of 5 officers and 9 civilians. If you must   Updated on: 26 Mar 2010, 06:54
1
Kudos
gmatinfoonline wrote:
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.

"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?"


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

Originally posted by bangalorian2000 on 24 Mar 2010, 09:47.
Last edited by bangalorian2000 on 26 Mar 2010, 06:54, edited 1 time in total.
avatar
sookol
Intern
Intern
Joined: 02 Mar 2010
Posts: 1
Own Kudos [?]: [0]
Given Kudos: 0
Send PM
Re: tricky combinatorics problem [#permalink] A group consists of 5 officers and 9 civilians. If you must   25 Mar 2010, 17:18
bangalorian2000 wrote:
total = 5C2 * 9C3 + 5C3 * 9C2 = 840 + 360 = 1170


a small mistake 840 + 360 = 1200



careless mistakes are sometimes the most painful
User avatar
nickk
Manager
Manager
Joined: 10 Aug 2009
Posts: 81
Own Kudos [?]: 39 [0]
Given Kudos: 13
Send PM
Re: tricky combinatorics problem [#permalink] A group consists of 5 officers and 9 civilians. If you must   26 Mar 2010, 04:51
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.
User avatar
L
stne
SVP
SVP
Joined: 27 May 2012
Posts: 1686
Own Kudos [?]: 1450 [0]
Given Kudos: 634
Send PM
Re: tricky combinatorics problem [#permalink] A group consists of 5 officers and 9 civilians. If you must   09 Sep 2012, 05:21
bangalorian2000 wrote:
gmatinfoonline wrote:
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.

"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?"


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


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?
User avatar
L
Bunuel
Math Expert
Joined: 02 Sep 2009
Posts: 93500
Own Kudos [?]: 627437 [3]
Given Kudos: 81978
Send PM
CAT Tests Forum Quiz Mode
Re: tricky combinatorics problem [#permalink] A group consists of 5 officers and 9 civilians. If you must   09 Sep 2012, 05:31
2
Kudos
1
Bookmarks
Expert Reply
stne wrote:
bangalorian2000 wrote:
gmatinfoonline wrote:
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.

"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?"


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


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?


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.
User avatar
B
TooLong150
Manager
Manager
Joined: 10 Mar 2013
Posts: 136
Own Kudos [?]: 495 [0]
Given Kudos: 2412
GMAT 1: 620 Q44 V31
GMAT 2: 610 Q47 V28
GMAT 3: 700 Q49 V36
GMAT 4: 690 Q48 V35
GMAT 5: 750 Q49 V42
GMAT 6: 730 Q50 V39
GPA: 3
Send PM
Reviews Badge
Re: A group consists of 5 officers and 9 civilians. If you must [#permalink] A group consists of 5 officers and 9 civilians. If you must   30 Oct 2013, 04:07
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?
User avatar
B
oss198
Manager
Manager
Joined: 18 Jul 2013
Posts: 69
Own Kudos [?]: 364 [0]
Given Kudos: 120
Location: Italy
Schools: LBS '21 Insead Sept19 Intake (A) IESE '21 (A$)
GMAT 1: 600 Q42 V31
GMAT 2: 700 Q48 V38
GPA: 3.75
Send PM
Re: tricky combinatorics problem [#permalink] A group consists of 5 officers and 9 civilians. If you must   13 Nov 2013, 16:29
Bunuel wrote:
Hi all,


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.


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?
avatar
B
oli29
Intern
Intern
Joined: 06 Jul 2017
Posts: 6
Own Kudos [?]: 0 [0]
Given Kudos: 22
Send PM
Re: A group consists of 5 officers and 9 civilians. If you must [#permalink] A group consists of 5 officers and 9 civilians. If you must   22 Sep 2017, 08:06
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?
User avatar
bumpbot
Non-Human User
Joined: 09 Sep 2013
Posts: 33254
Own Kudos [?]: 831 [0]
Given Kudos: 0
Send PM
Re: A group consists of 5 officers and 9 civilians. If you must [#permalink] A group consists of 5 officers and 9 civilians. If you must   21 Dec 2021, 15:30
Hello from the GMAT Club BumpBot!

Thanks to another GMAT Club member, I have just discovered this valuable topic, yet it had no discussion for over a year. I am now bumping it up - doing my job. I think you may find it valuable (esp those replies with Kudos).

Want to see all other topics I dig out? Follow me (click follow button on profile). You will receive a summary of all topics I bump in your profile area as well as via email.
GMAT Club Bot
gmatclubot
Re: A group consists of 5 officers and 9 civilians. If you must [#permalink]
21 Dec 2021, 15:30
Moderator:
Bunuel
Math Expert
93500 posts