It is currently 17 Oct 2017, 19:19

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.

Close

Request Expert Reply

Confirm Cancel

Events & Promotions

Events & Promotions in June
Open Detailed Calendar

Two couples and one single person are seated at random in a

  post reply Question banks Downloads My Bookmarks Reviews Important topics  
Author Message
Director
Director
User avatar
Joined: 05 May 2004
Posts: 574

Kudos [?]: 67 [0], given: 0

Location: San Jose, CA
Two couples and one single person are seated at random in a [#permalink]

Show Tags

New post 19 Sep 2004, 22:14
This topic is locked. If you want to discuss this question please re-post it in the respective forum.

Two couples and one single person are seated at random in a row of five chairs. What is the probability that neither of the couples sits together in adjacent chairs?

Kudos [?]: 67 [0], given: 0

Director
Director
avatar
Joined: 20 Jul 2004
Posts: 590

Kudos [?]: 156 [0], given: 0

 [#permalink]

Show Tags

New post 20 Sep 2004, 06:27
Pb that neither couple sit together
= 1 - Pb that couples always sit together
= 1 - 3!/5!
= 1 - 1/20
= 19/20

Kudos [?]: 156 [0], given: 0

Senior Manager
Senior Manager
User avatar
Joined: 06 Dec 2003
Posts: 366

Kudos [?]: 14 [0], given: 0

Location: India
 [#permalink]

Show Tags

New post 20 Sep 2004, 13:17
total no of arrangement - 5! / 2! * 2! = 30
Both couple sit together - 3! = 6

Couple - 1 sits together + Both couple sit together = 4! / 2! = 6
Couple - 2 sits together + Both couple sit together = 4! / 2! = 6

so no of cases, when neither couple will sit together = 30 - (6 + 6 + 6) = 12
So probability = 12/30 = 2/5

whats the OA, dear ?

Dharmin
_________________

Perseverance, Hard Work and self confidence

Kudos [?]: 14 [0], given: 0

GMAT Club Legend
GMAT Club Legend
avatar
Joined: 15 Dec 2003
Posts: 4285

Kudos [?]: 527 [0], given: 0

 [#permalink]

Show Tags

New post 20 Sep 2004, 14:56
I have 4/5
Total possible outcomes: 5!
unfavorable outcomes with couple XX, YY sitting together:
(XX)(YY)(Z)
Since Xs and Ys are interchangeable, we have 3!*2!*2!
1 - (3!*2!*2!)/5! = 96/120 = 4/5
hardworker, I think you forgot to add 2!*2! for when X's and Y's are interchanged
_________________

Best Regards,

Paul

Kudos [?]: 527 [0], given: 0

Senior Manager
Senior Manager
avatar
Joined: 19 May 2004
Posts: 291

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

 [#permalink]

Show Tags

New post 20 Sep 2004, 15:23
Paul, you have calculated the probability of 1-(BOTH are together).
There is also a possibility that only one of the couples sits together.

total = 5! = 120.
both together = 3!*2!*2! = 24.
one together = 4!*2 - 24 = 24

The answer should be 2/5

Last edited by Dookie on 20 Sep 2004, 15:47, edited 1 time in total.

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

GMAT Club Legend
GMAT Club Legend
avatar
Joined: 15 Dec 2003
Posts: 4285

Kudos [?]: 527 [0], given: 0

 [#permalink]

Show Tags

New post 20 Sep 2004, 15:39
:oops: True, true... Dharmin was right :good
_________________

Best Regards,

Paul

Kudos [?]: 527 [0], given: 0

Director
Director
avatar
Joined: 20 Jul 2004
Posts: 590

Kudos [?]: 156 [0], given: 0

 [#permalink]

Show Tags

New post 20 Sep 2004, 17:54
Thats why I shouldn't answer at 5:30 AM. :evil: Good question though.

Total number of cases = 5! = 120
Number of cases both couples sit together = 3! X 2 X 2 = 24
Number of cases first couple sits together = 4! X 2 = 48 - 24 = 24
Number of cases second couple sits together = 4! X 2 = 48 - 24 = 24

Pb that both couples do not sit together
= 1 - Pb that both sit together - Pb that first couple sits together - Pb that second couple sits together
= 1 - (24+24+24)/120
= 1 - 3/5
2/5

Kudos [?]: 156 [0], given: 0

Intern
Intern
avatar
Joined: 02 Aug 2004
Posts: 25

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

 [#permalink]

Show Tags

New post 21 Sep 2004, 16:40
Thanks for the heads up. I'll be more careful when i post i guess.

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

Senior Manager
Senior Manager
avatar
Joined: 25 Jul 2004
Posts: 272

Kudos [?]: 11 [0], given: 0

 [#permalink]

Show Tags

New post 21 Sep 2004, 18:08
Consider the placement of the single person:

Single in Chair 1:
4 possible for chair 2
2 possible for chair 1

8 arrangements
(same as if single in Chair 5)

Single in Chair 2:
4 possible for chair 1
2 possible for chair 3

8 Arrangements
(same as if single in Chair 4)

Single in the Middle
4 possible in chair 1
2 possible in chair 2
2 possible in chair 4
16 arangements

probablility = 8 + 8 + 16 + 8 + 8 / 5!
= 48/120 = 2/5

Kudos [?]: 11 [0], given: 0

  [#permalink] 21 Sep 2004, 18:08
Display posts from previous: Sort by

Two couples and one single person are seated at random in a

  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 | Emoji artwork provided by EmojiOne

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®.