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

It is currently 21 Oct 2014, 18:24

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

There are 4 married couples in the ball room, 4 people are

  Question banks Downloads My Bookmarks Reviews Important topics  
Author Message
TAGS:
Intern
Intern
avatar
Joined: 14 Apr 2006
Posts: 16
Location: london
Followers: 0

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

 [#permalink] New post 07 May 2006, 13:05
prude_sb wrote:
Yes 8/35

We need to select one person out of each couple (2C1) of the 4 couples

Total number of ways: 2^4

Total number of ways of selecting 4 ppl out of 8 ppl is 8C4


Probability = (2*2*2*2)*(4*3*2*1)/(8*7*6*5) = 8/35


Ok you are choosing 4 people one person from each of the four couples,but this is not correct because you can choose a married couple but only that they shouldnt be the dancing pair
Example you can choose
M1W1 W3 M4
but though you have chosen M1 W1
M1 dances with W3 and M4 dances with W1 so none of them is dancing with their own spouse
Manager
Manager
avatar
Joined: 20 Mar 2005
Posts: 201
Location: Colombia, South America
Followers: 1

Kudos [?]: 6 [0], given: 0

 [#permalink] New post 07 May 2006, 14:00
sorry no OA, I just made it up, but I think proffesor's answer is the correct one
Senior Manager
Senior Manager
avatar
Joined: 12 Mar 2006
Posts: 371
Schools: Kellogg School of Management
Followers: 2

Kudos [?]: 28 [0], given: 3

 [#permalink] New post 09 May 2006, 19:08
fighter wrote:
prude_sb wrote:
Yes 8/35

We need to select one person out of each couple (2C1) of the 4 couples

Total number of ways: 2^4

Total number of ways of selecting 4 ppl out of 8 ppl is 8C4


Probability = (2*2*2*2)*(4*3*2*1)/(8*7*6*5) = 8/35


Ok you are choosing 4 people one person from each of the four couples,but this is not correct because you can choose a married couple but only that they shouldnt be the dancing pair
Example you can choose
M1W1 W3 M4
but though you have chosen M1 W1
M1 dances with W3 and M4 dances with W1 so none of them is dancing with their own spouse




Yes you are right . Let me try again

Number of ways by which we can select 2 couples from 4 couples = 4C2

So the probability that the couples never dance together is
1 - 4C2/8C4 = 1 - 2*3/(2*7*5) = 66/70
VP
VP
User avatar
Joined: 29 Dec 2005
Posts: 1351
Followers: 7

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

 [#permalink] New post 09 May 2006, 19:16
conocieur wrote:
sorry no OA, I just made it up, but I think proffesor's answer is the correct one


:banana

after all i am your buddy :P :P :P
Senior Manager
Senior Manager
avatar
Joined: 29 Jun 2005
Posts: 403
Followers: 1

Kudos [?]: 15 [0], given: 0

 [#permalink] New post 10 May 2006, 23:24
Sorry, pushed submit twice...
Can't delete this post

Last edited by Dilshod on 10 May 2006, 23:44, edited 1 time in total.
Senior Manager
Senior Manager
avatar
Joined: 29 Jun 2005
Posts: 403
Followers: 1

Kudos [?]: 15 [0], given: 0

 [#permalink] New post 10 May 2006, 23:29
Hi,
ab cd ef gh suppose they are 4 married couples.
There are only 6 choices that assure that none of the chosen people are married: they are aceg, bdfh, bdeg, adeg, adfg, acfh
If a is chosen, then the second person should be c, e or h. The probability is 3/7.
If c is chosen, then the third person should be e or h. The probability is 2/6
If e is chosen, then the fourth person should be h. The probability is 1/5
Product of three probabilities = 2/7*1/3*1/5=2/105.
The same works for remaining combinations, so just multiply by 6 and get 12/105.
The answer is 4/35.
  [#permalink] 10 May 2006, 23:29
    Similar topics Author Replies Last post
Similar
Topics:
21 Experts publish their posts in the topic If 4 people are selected from a group of 6 married couples bibha 19 13 Aug 2010, 07:42
Six Married couples are standing in a room. If 4 people are neelesh 9 17 Dec 2007, 20:42
6 married couple are present at a party. If 4 people are shoonya 5 07 Aug 2007, 19:23
6 married couples are present at a party. if 4 people are shoonya 11 06 Aug 2006, 15:05
Six Married couples are standing in a room. If 4 people are neelesh 18 08 Jan 2005, 21:15
Display posts from previous: Sort by

There are 4 married couples in the ball room, 4 people are

  Question banks Downloads My Bookmarks Reviews Important topics  

Go to page   Previous    1   2   [ 26 posts ] 



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®.