It is currently 20 Jan 2018, 09:14

### 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

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

# GRE Math Challenge #13 - Six people are asked to sit down in

Author Message
TAGS:
CEO
Joined: 07 Jun 2014
Posts: 3018
GRE 1: 323 Q167 V156
WE: Business Development (Energy and Utilities)
Followers: 53

Kudos [?]: 873 [2] , given: 154

GRE Math Challenge #13 - Six people are asked to sit down in [#permalink]  19 Sep 2014, 02:39
2
KUDOS
Expert's post
00:00

Question Stats:

0% (00:00) correct 0% (00:00) wrong based on 0 sessions
Six people are asked to sit down in a circle consisting of eight chairs.Assume that the six people are in fact three couples.What is the probability that at least one of the three wives will sit next to her husband if everybody takes a seat randomly.

P = 6/7
Number of arrangements where at least one wife sits next to her husband:

_________________

Sandy
If you found this post useful, please let me know by pressing the Kudos Button

Try our free Online GRE Test

Intern
Joined: 28 Jun 2014
Posts: 6
Followers: 0

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

Re: GRE Math Challenge #13 [#permalink]  25 Sep 2014, 12:33
pls explain me ............
CEO
Joined: 07 Jun 2014
Posts: 3018
GRE 1: 323 Q167 V156
WE: Business Development (Energy and Utilities)
Followers: 53

Kudos [?]: 873 [1] , given: 154

Re: GRE Math Challenge #13 [#permalink]  26 Sep 2014, 01:38
1
KUDOS
Expert's post
The question states at least one partner sits next to other in the circular arrangement. Now Initially we had to arrange 8 objects in a circular formation. 6 persons(distinguishable) and two vacant spaces (non distinguishable). So number of arrangements would be 8!/8*2! = 2520.

Now lets assume we have a couple now the problem becomes 4 persons + 1 couple + 2 blank spaces in 7 chairs. So number of arrangements 7!/ 7 * 2! = 2160.

Divide to obtain probability: 6/7.
_________________

Sandy
If you found this post useful, please let me know by pressing the Kudos Button

Try our free Online GRE Test

Intern
Joined: 23 Jun 2017
Posts: 6
Followers: 0

Kudos [?]: 5 [1] , given: 4

Re: GRE Math Challenge #13 [#permalink]  02 Jul 2017, 10:54
1
KUDOS
sandy wrote:
The question states at least one partner sits next to other in the circular arrangement. Now Initially we had to arrange 8 objects in a circular formation. 6 persons(distinguishable) and two vacant spaces (non distinguishable). So number of arrangements would be 8!/8*2! = 2520.

Now lets assume we have a couple now the problem becomes 4 persons + 1 couple + 2 blank spaces in 7 chairs. So number of arrangements 7!/ 7 * 2! = 2160.

Divide to obtain probability: 6/7.

won't the two people who form the couple arrange among themselves in 2 ways?
Moderator
Joined: 18 Apr 2015
Posts: 2674
Followers: 39

Kudos [?]: 382 [0], given: 1744

Re: GRE Math Challenge #13 - Six people are asked to sit down in [#permalink]  03 Jul 2017, 01:26
Expert's post
Waiting the experts
_________________
Senior Manager
Joined: 20 Apr 2016
Posts: 322
Followers: 1

Kudos [?]: 201 [2] , given: 55

Re: GRE Math Challenge #13 - Six people are asked to sit down in [#permalink]  04 Jul 2017, 08:53
2
KUDOS
sandy wrote:
Six people are asked to sit down in a circle consisting of eight chairs.Assume that the six people are in fact three couples.What is the probability that at least one of the three wives will sit next to her husband if everybody takes a seat randomly.

P = 6/7
Number of arrangements where at least one wife sits next to her husband:

Using complement rule
That is, P(Event A happening) = 1 - P(Event A not happening)

So, here we get: P(at least 1 matching couple) = 1 - P(zero matching couple)

P(zero matching couple) =6/8 x 4/7 x 2/6
= 1/7

So, P(at least 1 couple) = 1 - P(no couple)
= 1 - 1/7
= 6/7
_________________

If you found this post useful, please let me know by pressing the Kudos Button

Re: GRE Math Challenge #13 - Six people are asked to sit down in   [#permalink] 04 Jul 2017, 08:53
Display posts from previous: Sort by