Last visit was: 13 May 2025, 01:01 It is currently 13 May 2025, 01:01
Decision Tracker
My Rewards
New comers' posts
New posts
Unanswered

Close

GRE Prep 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 GRE 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.

Go to My Error Log Learn more
Close

Request Expert Reply

Confirm Cancel
PREVIOUS TOPIC
NEXT TOPIC

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

Tags :
Find Similar Topics  About  Add a Tag

Show Tags
Hide Tags

User avatar
sandy
Retired Moderator
Joined: 07 Jun 2014
Posts: 4815
Own Kudos [?]: 11404 [24]
Given Kudos: 0
GRE 1: Q167 V156
WE:Business Development (Energy and Utilities)
Send PM
GRE Math Challenge #13 - Six people are asked to sit down in [#permalink] New post   19 Sep 2014, 02:39
5
Expert Reply
19
Bookmarks
00:00

Question Stats:

9% (01:26) correct 90% (02:44) wrong based on 43 sessions

Hide Show timer Statistics

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.

Show: :: Answer
P = 6/7
Number of arrangements where at least one wife sits next to her husband:
Most Helpful Expert Reply
User avatar
sandy
Retired Moderator
Joined: 07 Jun 2014
Posts: 4815
Own Kudos [?]: 11404 [5]
Given Kudos: 0
GRE 1: Q167 V156
WE:Business Development (Energy and Utilities)
Send PM
Re: GRE Math Challenge #13 [#permalink] New post   26 Sep 2014, 01:38
4
Expert Reply
1
Bookmarks
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.
Most Helpful Community Reply
avatar
pranab223
Retired Moderator
Joined: 20 Apr 2016
Posts: 1307
Own Kudos [?]: 2298 [9]
Given Kudos: 251
WE:Engineering (Energy and Utilities)
Send PM

Top Member of the Month
Re: GRE Math Challenge #13 - Six people are asked to sit down in [#permalink] New post   04 Jul 2017, 08:53
7
2
Bookmarks
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.

Show: :: Answer
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
General Discussion
avatar
onindkamboj
Intern
Intern
Joined: 28 Jun 2014
Posts: 6
Own Kudos [?]: 2 [1]
Given Kudos: 0
Send PM
Re: GRE Math Challenge #13 [#permalink] New post   25 Sep 2014, 12:33
1
pls explain me ............
avatar
AmanSiso
Intern
Intern
Joined: 23 Jun 2017
Posts: 5
Own Kudos [?]: 8 [2]
Given Kudos: 0
Send PM
Re: GRE Math Challenge #13 [#permalink] New post   02 Jul 2017, 10:54
2
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?
User avatar
D
Carcass
Verbal Expert
Joined: 18 Apr 2015
Posts: 31927
Own Kudos [?]: 38028 [0]
Given Kudos: 26286
Send PM
Re: GRE Math Challenge #13 - Six people are asked to sit down in [#permalink] New post   03 Jul 2017, 01:26
Expert Reply
Waiting the experts
User avatar
Farina
Senior Manager
Senior Manager
Joined: 10 Feb 2020
Posts: 496
Own Kudos [?]: 358 [0]
Given Kudos: 299
Send PM
Re: GRE Math Challenge #13 - Six people are asked to sit down in [#permalink] New post   04 Aug 2020, 06:53
I am unable to understand this :(
avatar
etunamach
Intern
Intern
Joined: 06 Oct 2020
Posts: 1
Own Kudos [?]: 6 [6]
Given Kudos: 0
Send PM
Re: GRE Math Challenge #13 - Six people are asked to sit down in [#permalink] New post   10 Nov 2020, 08:23
5
1
Bookmarks
I believe this could be solved very simply following this path:

probability that one wife will sit next to her husband is 1/7. Since the arrangement is circular, the husband can be on either side of his wife, so probability increases by 2: 1/7x2=2/7

Since there are three couples, probability increases by 3: 2/7x3=6/7

Hence, the answer is 6/7
avatar
abaskm
Intern
Intern
Joined: 25 Jan 2020
Posts: 29
Own Kudos [?]: 44 [0]
Given Kudos: 25
Send PM
Re: GRE Math Challenge #13 [#permalink] New post   20 Nov 2020, 06:55
sandy wrote:

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.



You forgot to multiply by 3.
avatar
S
void
Senior Manager
Senior Manager
Joined: 03 Dec 2020
Posts: 440
Own Kudos [?]: 62 [0]
Given Kudos: 68
Send PM
Re: GRE Math Challenge #13 - Six people are asked to sit down in [#permalink] New post   10 Dec 2020, 12:46
this type of problem come in gre exam??
User avatar
D
Carcass
Verbal Expert
Joined: 18 Apr 2015
Posts: 31927
Own Kudos [?]: 38028 [0]
Given Kudos: 26286
Send PM
Re: GRE Math Challenge #13 - Six people are asked to sit down in [#permalink] New post   10 Dec 2020, 13:09
Expert Reply
void wrote:
this type of problem come in gre exam??


Could be. It is a bit off but legit
avatar
S
Gocha
Manager
Manager
Joined: 25 Aug 2020
Posts: 80
Own Kudos [?]: 68 [3]
Given Kudos: 65
Send PM
Re: GRE Math Challenge #13 - Six people are asked to sit down in [#permalink] New post   20 Dec 2020, 15:20
3
pranab223 wrote:
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.

Show: :: Answer
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


Hi, Thank you for explanation, I am trying to understand how your equation meets which use cases.
It does not change conclusion, but, should' P(zero matching couple) be, = 5/7 x 3/5 x 1/3 = 1/7 ?

#1st couple, (F1,F2)
F1 seat can be anything from 8 empty seats.

Remaining 7 (8-1) seats. (?/7)

F2 cannot be next to F1, so need to be picked from 5 seats (7-2)
5/7

#2nd couple, (S1,S2)
S1 seat can be anything from remaining 6 seats. (6 = (8-2) as 2 seats are taken by F1,F2)

Remaining 5(6-1) seats. (?/5)

S2 cannot be next to S1, so need to be picked from 3 seats (5-2)
3/5

#3rd couple (T1, T2)
T1 seat can be anything from remaining 4 seats. (4 = (8-4) as 4 seats are taken by F1,F2, S2,S2)

Remaining 3 (4-1) seats (?/3)
T2 cannot be next to T1,so need to be picked from 1 seats (3-2)
1/3

5/7 * 3/5 * 1/3 = 1/7

I am trying to understand whether above equation means same concept which you explained.
Thank you in advance.
avatar
B
Ola1100
Intern
Intern
Joined: 05 Jan 2023
Posts: 22
Own Kudos [?]: 16 [0]
Given Kudos: 35
GRE 1: Q164 V160
Send PM
Re: GRE Math Challenge #13 - Six people are asked to sit down in [#permalink] New post   06 Jul 2023, 05:10
pranab223 wrote:
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.

Show: :: Answer
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



Please how did you get the 6/8 * 4/7 * 2/6?

Can you please explain?
User avatar
D
Carcass
Verbal Expert
Joined: 18 Apr 2015
Posts: 31927
Own Kudos [?]: 38028 [0]
Given Kudos: 26286
Send PM
Re: GRE Math Challenge #13 - Six people are asked to sit down in [#permalink] New post   06 Jul 2023, 06:10
Expert Reply
8 sits and 6 people and 3 couples

Pick one and you have 6 over 8 = 6/8

Pick the second and you have 2 couples over one less sit 4/7

Pick third and you have with the same logic 2/6

Multiply together and you have 1/7 which is the event that will NOT happen.

using 1-1/7= and you will have all the events that will happen 6/7
User avatar
bumpbot
GRE Prep Club Legend
GRE Prep Club Legend
Joined: 07 Jan 2021
Posts: 5303
Own Kudos [?]: 77 [0]
Given Kudos: 0
Send PM
Re: GRE Math Challenge #13 - Six people are asked to sit down in [#permalink] New post   01 Aug 2024, 06:21
Hello from the GRE Prep Club BumpBot!

Thanks to another GRE Prep Club member, I have just discovered this valuable topic, yet it had no discussion for over a year. I am now bumping it up - doing my job. I think you may find it valuable (esp those replies with Kudos).

Want to see all other topics I dig out? Follow me (click follow button on profile). You will receive a summary of all topics I bump in your profile area as well as via email.
Prep Club for GRE Bot
gmatclubot
Re: GRE Math Challenge #13 - Six people are asked to sit down in [#permalink]
01 Aug 2024, 06:21
Moderators:
adewale223
GRE Instructor
110 posts
bronaugust
GRE Forum Moderator
37 posts
BrushMyQuant
Moderator
1131 posts
VibhuAnurag
GRE Instructor
234 posts

Powered by phpBB © phpBB Group | Emoji artwork provided by EmojiOne

Main navigation

Partners

GRE Resources

www.ets.org/gre | About | Terms and Conditions and Privacy Policy | Prep Club for GRE Rules | Contact