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

It is currently 29 May 2016, 02:06
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 committee of 3 people is to be chosen from four married

  post reply Question banks Downloads My Bookmarks Reviews Important topics  
Author Message
Intern
Intern
avatar
Joined: 20 Apr 2006
Posts: 39
Followers: 0

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

A committee of 3 people is to be chosen from four married [#permalink]

Show Tags

New post 01 Aug 2006, 05:11
This topic is locked. If you want to discuss this question please re-post it in the respective forum.

A committee of 3 people is to be chosen from four married couple. What is the number of different committees that can be chosen if 2 people who are married to each other cannot both serve on the committee.

1. 16
2. 24
3. 26
4. 30.
5. 32
CEO
CEO
User avatar
Joined: 20 Nov 2005
Posts: 2911
Schools: Completed at SAID BUSINESS SCHOOL, OXFORD - Class of 2008
Followers: 21

Kudos [?]: 195 [0], given: 0

 [#permalink]

Show Tags

New post 01 Aug 2006, 09:25
E
Number of ways of choosing first person = 8
Number of ways of choosing second person = 6 (Out of 7 remaning, anyone can be chosen except his/her partner)
Number of ways of choosing third person = 4 (Out of 6 remaning anyone can be chosen except partners of two selected person)

Total = 8*6*4
But this contain duplicates. To remove duplicates divide by the factorial of number of people in the committee.
So total ways = 8*6*4/3! = 32
_________________

SAID BUSINESS SCHOOL, OXFORD - MBA CLASS OF 2008

Intern
Intern
avatar
Joined: 20 Apr 2006
Posts: 39
Followers: 0

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

 [#permalink]

Show Tags

New post 01 Aug 2006, 09:42
Thanks for the answer. Can this question also be solved using the xCn (combination) formula?
Intern
Intern
avatar
Joined: 02 May 2006
Posts: 25
Followers: 0

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

Re [#permalink]

Show Tags

New post 01 Aug 2006, 10:02
Yes, that is:

8C3-4*6C1

Where, 8C3 is the different ways you can choose 3 persons from the 8 possibilities.
PPX=groups where the partners are together..so that is 4*6=24

56-24=32
SVP
SVP
avatar
Joined: 30 Mar 2006
Posts: 1737
Followers: 1

Kudos [?]: 61 [0], given: 0

 [#permalink]

Show Tags

New post 02 Aug 2006, 03:51
E 32

Number of ways 3 people can be chosen from 8 people = 8C3 = 56

When the fist married couple is chosen .... number of ways to choose the third person = 6
Similarily since there are four couples.... number of ways to choose couple in the commitee = 4*6 = 24

Hence number of ways NOT to choose any couple = 56-24 = 32
  [#permalink] 02 Aug 2006, 03:51
Display posts from previous: Sort by

A committee of 3 people is to be chosen from four married

  post reply Question banks Downloads My Bookmarks Reviews Important topics  


GMAT Club MBA Forum Home| About| 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®.