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.

It appears that you are browsing the GMAT Club forum unregistered!

Signing up is free, quick, and confidential.
Join other 500,000 members and get the full benefits of GMAT Club

Registration gives you:

Tests

Take 11 tests and quizzes from GMAT Club and leading GMAT prep companies such as Manhattan GMAT,
Knewton, and others. All are free for GMAT Club members.

Applicant Stats

View detailed applicant stats such as GPA, GMAT score, work experience, location, application
status, and more

Books/Downloads

Download thousands of study notes,
question collections, GMAT Club’s
Grammar and Math books.
All are free!

Thank you for using the timer!
We noticed you are actually not timing your practice. Click the START button first next time you use the timer.
There are many benefits to timing your practice, including:

Juan and his five friends will sit on six fixed seats around [#permalink]

Show Tags

22 Sep 2009, 20:13

6

This post was BOOKMARKED

00:00

A

B

C

D

E

Difficulty:

35% (medium)

Question Stats:

62% (02:08) correct
38% (01:04) wrong based on 405 sessions

HideShow timer Statistics

Juan and his five friends will sit on six fixed seats around a circular table. If Juan must sit on the seat closest to the window and Jamal must sit next to Juan, in how many can Juan and his five friends sit?

count juan and jamal as one member that sits on a fixed chair. now we are left with 3 more members who can sit in 3! ways. jamal can sit on either side of juan so total combinations = 2*3! = 48

Juan and his five friends will sit on six fixed seats around a circular table. If Juan must sit on the seat closest to the window and Jamal must sit next to Juan, in how many can Juan and his five friends sit?

(A) 20 (B) 24 (C) 48 (D) 72 (E) 120

count juan and jamal as one member that sits on a fixed chair. now we are left with 3 more members who can sit in 3! ways. jamal can sit on either side of juan so total combinations = 2*3! = 48

ans: c

2*3! = 12 not 48, also there are total of 6 people so without Juan and Jamal there are 4 left not 3.

As the Juan's seat is fixed and Jamal must sit next to him then they can be seated only in 2 ways: {Juan}{Jamal} or {Jamal}{Juan}. Other 4 friends can be seated in 4! ways, so total 2*4!=48. _________________

Juan and his five friends will sit on six fixed seats around a circular table. If Juan must sit on the seat closest to the window and Jamal must sit next to Juan, in how many can Juan and his five friends sit?

(A) 20 (B) 24 (C) 48 (D) 72 (E) 120

count juan and jamal as one member that sits on a fixed chair. now we are left with 3 more members who can sit in 3! ways. jamal can sit on either side of juan so total combinations = 2*3! = 48

ans: c

2*3! = 12 not 48, also there are total of 6 people so without Juan and Jamal there are 4 left not 3.

As the Juan's seat is fixed and Jamal must sit next to him then they can be seated only in 2 ways: {Juan}{Jamal} or {Jamal}{Juan}. Other 4 friends can be seated in 4! ways, so total 2*4!=48.

Isn't the result 24?! Juan must sit on the seat closest to the window and cannot swap his seat with Jamal because if they swap seats Juan won't be closest to the window (which is a must). The others can sit in 4! ways. _________________

Juan and his five friends will sit on six fixed seats around a circular table. If Juan must sit on the seat closest to the window and Jamal must sit next to Juan, in how many can Juan and his five friends sit?

(A) 20 (B) 24 (C) 48 (D) 72 (E) 120

count juan and jamal as one member that sits on a fixed chair. now we are left with 3 more members who can sit in 3! ways. jamal can sit on either side of juan so total combinations = 2*3! = 48

ans: c

2*3! = 12 not 48, also there are total of 6 people so without Juan and Jamal there are 4 left not 3.

As the Juan's seat is fixed and Jamal must sit next to him then they can be seated only in 2 ways: {Juan}{Jamal} or {Jamal}{Juan}. Other 4 friends can be seated in 4! ways, so total 2*4!=48.

Isn't the result 24?! Juan must sit on the seat closest to the window and cannot swap his seat with Jamal because if they swap seats Juan won't be closest to the window (which is a must). The others can sit in 4! ways.

Juan's seat is fixed, yes. But Jamal can be to the right of him or to the left. _________________

What about the fact that the table is circular? Don't we need to use the formula (n-1)! say instead of 4! we would have 3!?

Thanks

Cheers! J

Since Juan's seat is fixed, then circular arrangement is not applicable here. You see, since Juan's seat is fixed, the number of arrangements of four friends (except Jamal) is 4! irrespective whether they are in front of Juan or in circle with him.

consider Juan and Jamal as one unit (as they always sit at adjacent places), as if we had 5 people altogether. using the formula for circular arrangement (n-1)! we have (5-1)!, and then multiplying the result be 2! to take into account arrangements inside our unit Juan+Jamal

Juan and his five friends will sit on six fixed seats around a circular table. If Juan must sit on the seat closest to the window and Jamal must sit next to Juan, in how many can Juan and his five friends sit?

(A) 20 (B) 24 (C) 48 (D) 72 (E) 120

count juan and jamal as one member that sits on a fixed chair. now we are left with 3 more members who can sit in 3! ways. jamal can sit on either side of juan so total combinations = 2*3! = 48

ans: c

2*3! = 12 not 48, also there are total of 6 people so without Juan and Jamal there are 4 left not 3.

As the Juan's seat is fixed and Jamal must sit next to him then they can be seated only in 2 ways: {Juan}{Jamal} or {Jamal}{Juan}. Other 4 friends can be seated in 4! ways, so total 2*4!=48.

Hi Bunuel, could you please explain why do we multiply by 2 in the following 2*4! (its 4! because we are considering Juan and jamal as 1 so thats fine).

Juan and his five friends will sit on six fixed seats around a circular table. If Juan must sit on the seat closest to the window and Jamal must sit next to Juan, in how many can Juan and his five friends sit?

(A) 20 (B) 24 (C) 48 (D) 72 (E) 120

count juan and jamal as one member that sits on a fixed chair. now we are left with 3 more members who can sit in 3! ways. jamal can sit on either side of juan so total combinations = 2*3! = 48

ans: c

2*3! = 12 not 48, also there are total of 6 people so without Juan and Jamal there are 4 left not 3.

As the Juan's seat is fixed and Jamal must sit next to him then they can be seated only in 2 ways: {Juan}{Jamal} or {Jamal}{Juan}. Other 4 friends can be seated in 4! ways, so total 2*4!=48.

Hi Bunuel, could you please explain why do we multiply by 2 in the following 2*4! (its 4! because we are considering Juan and jamal as 1 so thats fine).

Really bad with combinations and probability

Those two cane be arranged either {Juan}{Jamal} or {Jamal}{Juan}. For each of these arrangements the remaining 4 can be seated in 4!, so total is 2*4!.

Juan and his five friends will sit on six fixed seats around a circular table. If Juan must sit on the seat closest to the window and Jamal must sit next to Juan, in how many can Juan and his five friends sit?

(A) 20 (B) 24 (C) 48 (D) 72 (E) 120

count juan and jamal as one member that sits on a fixed chair. now we are left with 3 more members who can sit in 3! ways. jamal can sit on either side of juan so total combinations = 2*3! = 48

ans: c

2*3! = 12 not 48, also there are total of 6 people so without Juan and Jamal there are 4 left not 3.

As the Juan's seat is fixed and Jamal must sit next to him then they can be seated only in 2 ways: {Juan}{Jamal} or {Jamal}{Juan}. Other 4 friends can be seated in 4! ways, so total 2*4!=48.

Hi Bunuel, could you please explain why do we multiply by 2 in the following 2*4! (its 4! because we are considering Juan and jamal as 1 so thats fine).

Really bad with combinations and probability

Those two cane be arranged either {Juan}{Jamal} or {Jamal}{Juan}. For each of these arrangements the remaining 4 can be seated in 4!, so total is 2*4!.

Re: Juan and his five friends will sit on six fixed seats around [#permalink]

Show Tags

17 Jun 2015, 08:29

Hello from the GMAT Club BumpBot!

Thanks to another GMAT 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. _________________

Re: Juan and his five friends will sit on six fixed seats around [#permalink]

Show Tags

25 Dec 2015, 13:24

Using FCP rule

seat1 - 1 way - John can sit at any place in 1 way Seat2 - 2 ways - Jamal should sit next to him. either left or right seat 3 - 4 ways - 4 people remaining. anyone can sit anywhere Seat 4 - 3 ways - 3 people can sit anywhere Seat 5 - 2 ways - 2 people can sit anywhere Finally, Seat6 - 1 way - 1 person & 1 seat available

Total = 1 x 1 x 4 x 3 x 2 x 1 = 48 ways

gmatclubot

Re: Juan and his five friends will sit on six fixed seats around
[#permalink]
25 Dec 2015, 13:24

Excellent posts dLo saw your blog too..!! Man .. you have got some writing skills. And Just to make an argument = You had such an amazing resume ; i am glad...

So Much $$$ Business school costs a lot. This is obvious, whether you are a full-ride scholarship student or are paying fully out-of-pocket. Aside from the (constantly rising)...

I barely remember taking decent rest in the last 60 hours. It’s been relentless with submissions, birthday celebration, exams, vacating the flat, meeting people before leaving and of...

Rishabh from Gyan one services, India had a one to one interview with me where I shared my experience at IMD till now. http://www.gyanone.com/blog/life-at-imd-interview-with-imd-mba/ ...