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

It is currently 31 Jul 2015, 05:47
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

I got it right but that was just a wild guess

  Question banks Downloads My Bookmarks Reviews Important topics  
Author Message
Manager
Manager
avatar
Joined: 27 Mar 2008
Posts: 72
Followers: 1

Kudos [?]: 8 [0], given: 0

I got it right but that was just a wild guess [#permalink] New post 09 Aug 2008, 17:56
This topic is locked. If you want to discuss this question please re-post it in the respective forum.

I got it right but that was just a wild guess.
Attachments

gmat ps3.jpg
gmat ps3.jpg [ 21.29 KiB | Viewed 762 times ]

Manager
Manager
avatar
Joined: 15 Jul 2008
Posts: 207
Followers: 3

Kudos [?]: 33 [0], given: 0

Re: Counting and Permutation and Probablity [#permalink] New post 09 Aug 2008, 18:18
How many ways can you choose the first member of the committee ? 8
How many ways can you choose the second member ? 6 (you are not allowed to choose the spouse of the first)
How many ways can you choose the third member ? 4 (spouses of the first two are not allowed)

so 8x6x4. But in this you have included the order of arrangements. 3x2x1 possible arrangements of these 3 members.

so your answer is 8x6x4 / 3! = 32
Director
Director
User avatar
Joined: 12 Jul 2008
Posts: 518
Schools: Wharton
Followers: 17

Kudos [?]: 124 [0], given: 0

Re: Counting and Permutation and Probablity [#permalink] New post 10 Aug 2008, 21:51
dishant007 wrote:
I got it right but that was just a wild guess.


Total ways of choosing 3 people = 8 choose 3 = 56
Total # of groups with a married couple = 4*6 = 24

Total # of groups w/o a married couple = 56-24 = 32
Manager
Manager
User avatar
Joined: 30 Jul 2007
Posts: 129
Followers: 2

Kudos [?]: 12 [0], given: 0

Re: Counting and Permutation and Probablity [#permalink] New post 11 Aug 2008, 12:46
zoinnk wrote:
dishant007 wrote:
I got it right but that was just a wild guess.


Total ways of choosing 3 people = 8 choose 3 = 56
Total # of groups with a married couple = 4*6 = 24

Total # of groups w/o a married couple = 56-24 = 32



Where does the 6 come from?
SVP
SVP
User avatar
Joined: 07 Nov 2007
Posts: 1824
Location: New York
Followers: 29

Kudos [?]: 572 [0], given: 5

Re: Counting and Permutation and Probablity [#permalink] New post 11 Aug 2008, 12:59
dishant007 wrote:
I got it right but that was just a wild guess.


No of ways form commite of 3 out of 8 members = 8C3 = 56

M1 M2 M3 M4 W1 W2 W3 W4

No of way for commite of 3 out of 8 with couples to gether
= Chose one man out of Four * Chose second member (man's spouse) * Chose third member (other than couple)
= 4C1 *1C1* 6C1
= 4 * 6 =24

Ans = 56-24=36
_________________

Your attitude determines your altitude
Smiling wins more friends than frowning

Re: Counting and Permutation and Probablity   [#permalink] 11 Aug 2008, 12:59
Display posts from previous: Sort by

I got it right but that was just a wild guess

  Question banks Downloads My Bookmarks Reviews Important topics  


cron

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®.