Last visit was: 29 Apr 2026, 23:40 It is currently 29 Apr 2026, 23:40
Home
Messages
Tests
Schools
Events
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.
Go to My Error Log Learn more
Close
Request Expert Reply
Confirm Cancel
Back to Forum
Create Topic
   1   2 
Poor Quality|      
User avatar
KarishmaB
User avatar
Joined: 16 Oct 2010
Last visit: 29 Apr 2026
Posts: 16,447
Own Kudos:
79,447
Given Kudos: 485
Location: Pune, India
Products:
GMAT Club Tests
Expert
Expert reply
Active GMAT Club Expert! Tag them with @ followed by their username for a faster response.
Posts: 16,447
Kudos: 79,447
There are 6 people at a party sitting at a round table with 6 seats: A 03 Jun 2019, 01:39
Kudos
Add Kudos
Bookmarks
Bookmark this Post
rishabhjain13
VeritasKarishma
mrdanielkim
i figured it out after taking a break from studying. thanks anyway!

here's the solution anyway, though the book has the question worded differently:

Q: IN how many ways can 6 people be seated at a round table if one of those seated cannot sit next to two of the other five:

A: Six people can be seated around a round table in 5! ways. there are 2 ways that the two unwelcome people could sit next to the person in question and 3! ways of arranging the other three. This is subtracted from the base case of 5!, giving the result of 108.

This question is different from the one posted above.

There are two versions and the answer would be different in the two cases.

Let me pick this version first:
There are 6 people (say A, B, C, D, E and F). They have to sit around a circular table such that one of them, say A, cannot sit next to D and F at the same time. (This means that A can sit next to D but not while F is on A's other side. Similarly, A can sit next to F too but not while D is on A's other side)

Total number of ways of arranging 6 people in a circle = 5! = 120

In how many of these 120 ways will A be between D and F?

We make DAF sit on three consecutive seats and make other 3 people sit in 3! ways.
or we make FAD sit of three consecutive seats and make other 3 people sit in 3! ways.
In all, we make A sit next to D and F simultaneously in 12 ways.

120 - 12 = 108 is the number of ways in which D and F are not sitting next to A at the same time.


The second version which seemed like the intended meaning of the original poster:
There are 6 people (say A, B, C, D, E and F). They have to sit around a circular table such that one of them, say A, can sit neither next to D nor next to F. (This means that A cannot sit next to D in any case and A cannot sit next to F in any case.)

Here, we say that A has to sit next to two of B, C and E.
Let's choose 2 of B, C and E in 3C2 = 3 ways. Let's arrange them around A in 2 ways (say we choose B and C. We could have BAC or CAB). We make these 3 sit on any 3 consecutive seats in 1 way. Number of ways of arranging these 3 people = 3*2 = 6

The rest of the 3 people can sit in 3! = 6 ways
Total number of ways in which A will sit neither next to D nor next to F = 6*6 = 36 ways

Hi VeritasKarishma

I used another method learnt from one of your posts itself, to solve this question. Please see if the application is correct.

If A can sit neither next to D nor next to F, I calculated the total cases (5!) from which I subtracted all the cases where A and D would necessarily sit together (2*4!) and the ones where A and F would necessarily sit together (2*4!). Now, here, there would be a slight overlap where A would be sitting between D and F (DAF and FAD), so adding that to the equation to remove the overlap (2*3!)

5! - (2*4! + 2*4! - 2*3!)
5! - 2*4! - 2*4! + 2*3!
120 - 48 - 48 + 12
=36 ways
Show more

Yes, you are using the sets concept here to solve the question. It is perfectly correct.
avatar
akdebonair
avatar
Joined: 09 Mar 2019
Last visit: 11 Mar 2021
Posts: 5
Own Kudos:
7
Given Kudos: 32
Posts: 5
Kudos: 7
There are 6 people at a party sitting at a round table with 6 seats: A 03 Jun 2019, 05:48
Kudos
Add Kudos
Bookmarks
Bookmark this Post
Selecting 1st seat for A : No. of available seat = 6 No. of ways to select 1 seat for A = 6
Now as we have selected one seat we can no longer apply circular arrangement formula.
Now selecting seat for D and F : No. of available seat = 3 (As we can not place them on either side of A) No. Of ways to select 2 seat out of 3 for D and F = 3
Now arranging B C and E in remaining 3 seats = 3!
Total Nos of possible arrangements = 6*3*3! = 108

( Experts kindly review the approach is it correct?)
User avatar
axezcole
User avatar
Joined: 25 Apr 2018
Last visit: 19 Apr 2022
Posts: 49
Own Kudos:
332
Given Kudos: 48
Location: India
Concentration: Finance, Technology
Schools: IIMU PGPX (A)
GMAT 1: 600 Q47 V25
Products:
My Reviews
Schools: IIMU PGPX (A)
GMAT 1: 600 Q47 V25
Posts: 49
Kudos: 332
There are 6 people at a party sitting at a round table with 6 seats: A 22 Oct 2019, 04:07
Kudos
Add Kudos
Bookmarks
Bookmark this Post
Another way -

I call it imagine people entering a meeting room one by one/VeritasKarishma's method.
Refer-https://gmatclub.com/forum/combinatorics-made-easy-206266.html
Topic name-circular arrangements

We have been said that A can sit neither next to D nor next to F.
So our answer is total ways- ways in which A can sit next to D or next to F.

Three conditions are possible-

A sits between D& F =1*2*1*3! (A comes in 1st and has 1 way to decide,D comes in 2nd and has 2 ways to decide,F comes in 3rd and has 1 way to decide,Rest have 3 ! ways to decide)

A sits only next to D(remember F cannot be adjacent to A here)=1*2*3*3! (A comes in 1st and has 1 way to decide,D comes in 2nd and has 2 ways to decide,F comes in 3rd and has 3 ways to decide,Rest have 3 ! ways to decide)

A sits only next to F(remember D cannot be adjacent to A)=1*2*3*3! (A comes in 1st and has 1 way to decide,F comes in 2nd and has 2 ways to decide,D comes in 3rd and has 3 ways to decide,Rest have 3 ! ways to decide)

Final answer-
1*2*3!+ 1*2*3*3!+ 1*2*3*3!=84


5!-84=36

Solve using a circular table diagram for the three conditions mentioned,it will make more sense.
User avatar
gvvsnraju@12
User avatar
Joined: 26 Sep 2016
Last visit: 28 Oct 2020
Posts: 29
Own Kudos:
35
Given Kudos: 25
Location: India
Schools: S.P.Jain Institute (A)
GMAT 1: 640 Q49 V27 (Online)
Schools: S.P.Jain Institute (A)
GMAT 1: 640 Q49 V27 (Online)
Posts: 29
Kudos: 35
There are 6 people at a party sitting at a round table with 6 seats: A 24 Nov 2019, 12:40
Kudos
Add Kudos
Bookmarks
Bookmark this Post
VeritasKarishma

Hi

i m completely confused to infer the exact meaning of constraints of following questions.

1. There are 6 people, A, B, C, D, E and F. They have to sit around a circular
table such that A cannot sit next to D and F at the same time. How many such
arrangements are possible?

2. There are 6 people, A, B, C, D, E and F. They have to sit around a circular
table such that A can sit neither next to D nor next to F. How many such
arrangements are possible?

Kindly help me in getting over the confusion....

Per my understanding, the solution for both these questions are same.
But Per OA & your Combinatorics article, solutions are different as the constraints given are completely diff.

Please elaborate with an example.

Thanks in advance

Raju
User avatar
KarishmaB
User avatar
Joined: 16 Oct 2010
Last visit: 29 Apr 2026
Posts: 16,447
Own Kudos:
79,447
 [1]
Given Kudos: 485
Location: Pune, India
Products:
GMAT Club Tests
Expert
Expert reply
Active GMAT Club Expert! Tag them with @ followed by their username for a faster response.
Posts: 16,447
Kudos: 79,447
 [1]
There are 6 people at a party sitting at a round table with 6 seats: A 24 Nov 2019, 23:46
1
Kudos
Add Kudos
Bookmarks
Bookmark this Post
gvvsnraju@12
VeritasKarishma

Hi

i m completely confused to infer the exact meaning of constraints of following questions.

1. There are 6 people, A, B, C, D, E and F. They have to sit around a circular
table such that A cannot sit next to D and F at the same time. How many such
arrangements are possible?

2. There are 6 people, A, B, C, D, E and F. They have to sit around a circular
table such that A can sit neither next to D nor next to F. How many such
arrangements are possible?

Kindly help me in getting over the confusion....

Per my understanding, the solution for both these questions are same.
But Per OA & your Combinatorics article, solutions are different as the constraints given are completely diff.

Please elaborate with an example.

Thanks in advance

Raju
Show more

Yes, they are different questions.

"A cannot sit next to D and F at the same time"
implies DAF is not possible and FAD is not possible. But AD (or DA) can sit together and FA (or AF) can sit together. Just that both F and D cannot sit next to A at the same time but one of them can sit next to A if the other does not.


"A can sit neither next to D nor next to F"
DA or AD is not possible. Also, AF or FA is not possible. Then of course, FAD or DAF is also not possible.
A cannot sit next to either of the two independently also.
User avatar
DimitriosL
User avatar
Joined: 03 Nov 2020
Last visit: 15 Mar 2021
Posts: 1
Given Kudos: 19
Posts: 1
Kudos: 0
There are 6 people at a party sitting at a round table with 6 seats: A 14 Nov 2020, 08:33
Kudos
Add Kudos
Bookmarks
Bookmark this Post
VeritasKarishma
mrdanielkim
i figured it out after taking a break from studying. thanks anyway!

here's the solution anyway, though the book has the question worded differently:

Q: IN how many ways can 6 people be seated at a round table if one of those seated cannot sit next to two of the other five:

A: Six people can be seated around a round table in 5! ways. there are 2 ways that the two unwelcome people could sit next to the person in question and 3! ways of arranging the other three. This is subtracted from the base case of 5!, giving the result of 108.

This question is different from the one posted above.

There are two versions and the answer would be different in the two cases.

Let me pick this version first:
There are 6 people (say A, B, C, D, E and F). They have to sit around a circular table such that one of them, say A, cannot sit next to D and F at the same time. (This means that A can sit next to D but not while F is on A's other side. Similarly, A can sit next to F too but not while D is on A's other side)

Total number of ways of arranging 6 people in a circle = 5! = 120

In how many of these 120 ways will A be between D and F?

We make DAF sit on three consecutive seats and make other 3 people sit in 3! ways.
or we make FAD sit of three consecutive seats and make other 3 people sit in 3! ways.
In all, we make A sit next to D and F simultaneously in 12 ways.

120 - 12 = 108 is the number of ways in which D and F are not sitting next to A at the same time.


The second version which seemed like the intended meaning of the original poster:
There are 6 people (say A, B, C, D, E and F). They have to sit around a circular table such that one of them, say A, can sit neither next to D nor next to F. (This means that A cannot sit next to D in any case and A cannot sit next to F in any case.)

Here, we say that A has to sit next to two of B, C and E.
Let's choose 2 of B, C and E in 3C2 = 3 ways. Let's arrange them around A in 2 ways (say we choose B and C. We could have BAC or CAB). We make these 3 sit on any 3 consecutive seats in 1 way. Number of ways of arranging these 3 people = 3*2 = 6

The rest of the 3 people can sit in 3! = 6 ways
Total number of ways in which A will sit neither next to D nor next to F = 6*6 = 36 ways
Show more

Karishma, would it be ok to calc the number of wayes of i.e AD, then the n.of Ways of say FAD and then substract them all from 5! ?

For example it could look smth like this = 5! - 2*4!- 6*3! which yields 36

This Question is Locked Due to Poor Quality
Hi there,
The question you've reached has been archived due to not meeting our community quality standards. No more replies are possible here.
Looking for better-quality questions? Check out the 'Similar Questions' block below for a list of similar but high-quality questions.
Want to join other relevant Problem Solving discussions? Visit our Problem Solving (PS) Forum for the most recent and top-quality discussions.
Thank you for understanding, and happy exploring!
   1   2 
Moderators:
Bunuel
Math Expert
109994 posts
Krunaal
Tuck School Moderator
852 posts