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Juan and his five friends will sit on six fixed seats around [#permalink]
22 Sep 2009, 19:13

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Difficulty:

35% (medium)

Question Stats:

62% (02:25) correct
38% (01:08) wrong based on 110 sessions

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?

Re: GMAT Combinatorics 4 [#permalink]
26 Feb 2012, 13:17

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

Re: GMAT Combinatorics 4 [#permalink]
26 Feb 2012, 13:27

2

This post received KUDOS

Expert's post

MBAhereIcome wrote:

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. _________________

Re: GMAT Combinatorics 4 [#permalink]
26 Feb 2012, 16:43

Bunuel wrote:

MBAhereIcome wrote:

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. _________________

Re: GMAT Combinatorics 4 [#permalink]
26 Feb 2012, 16:46

Expert's post

Stiv wrote:

Bunuel wrote:

MBAhereIcome wrote:

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. _________________

Re: GMAT Combinatorics 4 [#permalink]
11 Dec 2013, 01:48

1

This post received KUDOS

Expert's post

jlgdr wrote:

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.

Re: GMAT Combinatorics 4 [#permalink]
12 Dec 2013, 04:20

answering the question about (n-1)! formula

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

Re: GMAT Combinatorics 4 [#permalink]
08 May 2014, 05:12

Bunuel wrote:

MBAhereIcome wrote:

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).

Re: GMAT Combinatorics 4 [#permalink]
08 May 2014, 06:40

Expert's post

gauravsoni wrote:

Bunuel wrote:

MBAhereIcome wrote:

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: GMAT Combinatorics 4 [#permalink]
08 May 2014, 10:28

gauravsoni wrote:

Bunuel wrote:

MBAhereIcome wrote:

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!.