Find all School-related info fast with the new School-Specific MBA Forum

It is currently 31 Aug 2015, 01:24
GMAT Club Tests

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.

Events & Promotions

Events & Promotions in June
Open Detailed Calendar

a 5 member committe is to be formed from a group of 5

  Question banks Downloads My Bookmarks Reviews Important topics  
Author Message
Intern
Intern
avatar
Joined: 05 Aug 2008
Posts: 14
Followers: 0

Kudos [?]: 0 [0], given: 0

a 5 member committe is to be formed from a group of 5 [#permalink] New post 19 Aug 2008, 07:45
This topic is locked. If you want to discuss this question please re-post it in the respective forum.

a 5 member committe is to be formed from a group of 5 military officers and 9 civilians. if the committee must include at least 2 civilians and 2 military officers, in how many different ways can the committee be chosen?

119

1200

3240

3600

14,400
Senior Manager
Senior Manager
avatar
Joined: 06 Apr 2008
Posts: 449
Followers: 1

Kudos [?]: 75 [0], given: 1

Re: combinations [#permalink] New post 19 Aug 2008, 07:53
pattydread wrote:
a 5 member committe is to be formed from a group of 5 military officers and 9 civilians. if the committee must include at least 2 civilians and 2 military officers, in how many different ways can the committee be chosen?

119

1200

3240

3600

14,400


Two combos = (3 civlians , 2 millitary) + ( 2 civilians, 3 military) = 9C3 * 5C2 + 9C2 * 5C3 = 84*10 + 36*10 = 1200
Manager
Manager
avatar
Joined: 11 Apr 2008
Posts: 202
Followers: 2

Kudos [?]: 17 [0], given: 1

Re: combinations [#permalink] New post 19 Aug 2008, 12:37
There are two combinations here in a 5 member committee.
2 civilians 3 military officers and 3 civilians and 2 military officers.
For the case I- the possible number of combinations 5C2*9C3=10*84=840
For case II- the possible number of combinations 5C3*9C2=10*36=360
Total number of combinations 840+360=1200
_________________

Nobody dies a virgin, life screws us all.

Manager
Manager
avatar
Joined: 14 Jun 2008
Posts: 161
Followers: 1

Kudos [?]: 28 [0], given: 0

Re: combinations [#permalink] New post 19 Aug 2008, 12:43
pattydread wrote:
a 5 member committe is to be formed from a group of 5 military officers and 9 civilians. if the committee must include at least 2 civilians and 2 military officers, in how many different ways can the committee be chosen?

119

1200

3240

3600

14,400


isnt it
First choose 2/9 civilians * choose 2/5 officers * the remaining
= 9C2 * 5C2 * (7+3)
= (9 * 8)/2 * (5*4)/2 * 10
= 3600
Director
Director
User avatar
Joined: 12 Jul 2008
Posts: 518
Schools: Wharton
Followers: 18

Kudos [?]: 127 [0], given: 0

Re: combinations [#permalink] New post 19 Aug 2008, 13:11
pattydread wrote:
a 5 member committe is to be formed from a group of 5 military officers and 9 civilians. if the committee must include at least 2 civilians and 2 military officers, in how many different ways can the committee be chosen?

119

1200

3240

3600

14,400


D

(5 choose 2)*(9 choose 2)*10 = 3600
Manager
Manager
avatar
Joined: 15 Jul 2008
Posts: 207
Followers: 3

Kudos [?]: 37 [0], given: 0

Re: combinations [#permalink] New post 19 Aug 2008, 13:57
sset009 wrote:

isnt it
First choose 2/9 civilians * choose 2/5 officers * the remaining
= 9C2 * 5C2 * (7+3)
= (9 * 8)/2 * (5*4)/2 * 10
= 3600


No. In your approach you are triple couting all the options. So you need to divide by 3 to get the correct answer.

Let us say m1-m5 are the 5 military guys and c1-c9 are the 9 civvis. In your expression,
m1 m2 + c1 c2 + m3 group is counted as being different from m1 m3 + c1 c2 + m2 and m2 m3 + c1 c2 + m1[/. In actuality they are all same m1m2m3 and c1c2.
So for each case you are taking 3 cases. Divide by 3 adn you will get the right answer.
Re: combinations   [#permalink] 19 Aug 2008, 13:57
Display posts from previous: Sort by

a 5 member committe is to be formed from a group of 5

  Question banks Downloads My Bookmarks Reviews Important topics  


GMAT Club MBA Forum Home| About| Privacy Policy| Terms and Conditions| GMAT Club Rules| Contact| Sitemap

Powered by phpBB © phpBB Group and phpBB SEO

Kindly note that the GMAT® test is a registered trademark of the Graduate Management Admission Council®, and this site has neither been reviewed nor endorsed by GMAC®.