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Juan and his five friends will sit on six fixed seats around

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

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Re: GMAT Combinatorics 4 [#permalink] New post 22 Sep 2009, 20:13
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J = Juan, F = Jamal
Since J is always fixed, set J, set F relative to J, then see how many options there are:
J F 4 3 2 1 = 24 or
F J 4 3 2 1 = 24

24+24=48 … C
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Re: GMAT Combinatorics 4 [#permalink] New post 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

ans: c
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Re: GMAT Combinatorics 4 [#permalink] New post 26 Feb 2012, 13:27
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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.
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Re: GMAT Combinatorics 4 [#permalink] New post 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.
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Re: GMAT Combinatorics 4 [#permalink] New post 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.
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COLLECTION OF QUESTIONS:
PS: 1. Tough and Tricky questions; 2. Hard questions; 3. Hard questions part 2; 4. Standard deviation; 5. Tough Problem Solving Questions With Solutions; 6. Probability and Combinations Questions With Solutions; 7 Tough and tricky exponents and roots questions; 8 12 Easy Pieces (or not?); 9 Bakers' Dozen; 10 Algebra set. ,11 Mixed Questions, 12 Fresh Meat

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Re: GMAT Combinatorics 4 [#permalink] New post 10 Dec 2013, 13:08
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 :)
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Re: GMAT Combinatorics 4 [#permalink] New post 11 Dec 2013, 01:48
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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.

Hope it's clear.
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COLLECTION OF QUESTIONS:
PS: 1. Tough and Tricky questions; 2. Hard questions; 3. Hard questions part 2; 4. Standard deviation; 5. Tough Problem Solving Questions With Solutions; 6. Probability and Combinations Questions With Solutions; 7 Tough and tricky exponents and roots questions; 8 12 Easy Pieces (or not?); 9 Bakers' Dozen; 10 Algebra set. ,11 Mixed Questions, 12 Fresh Meat

DS: 1. DS tough questions; 2. DS tough questions part 2; 3. DS tough questions part 3; 4. DS Standard deviation; 5. Inequalities; 6. 700+ GMAT Data Sufficiency Questions With Explanations; 7 Tough and tricky exponents and roots questions; 8 The Discreet Charm of the DS ; 9 Devil's Dozen!!!; 10 Number Properties set., 11 New DS set.


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Re: GMAT Combinatorics 4 [#permalink] New post 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
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Re: GMAT Combinatorics 4 [#permalink] New post 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).

Really bad with combinations and probability :(
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Re: GMAT Combinatorics 4 [#permalink] New post 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!.

Does this make sense?

Theory on Combinations: math-combinatorics-87345.html

DS questions on Combinations: search.php?search_id=tag&tag_id=31
PS questions on Combinations: search.php?search_id=tag&tag_id=52

Tough and tricky questions on Combinations: hardest-area-questions-probability-and-combinations-101361.html

Theory on probability problems: math-probability-87244.html

All DS probability problems to practice: search.php?search_id=tag&tag_id=33
All PS probability problems to practice: search.php?search_id=tag&tag_id=54

_________________

NEW TO MATH FORUM? PLEASE READ THIS: ALL YOU NEED FOR QUANT!!!

PLEASE READ AND FOLLOW: 11 Rules for Posting!!!

RESOURCES: [GMAT MATH BOOK]; 1. Triangles; 2. Polygons; 3. Coordinate Geometry; 4. Factorials; 5. Circles; 6. Number Theory; 7. Remainders; 8. Overlapping Sets; 9. PDF of Math Book; 10. Remainders; 11. GMAT Prep Software Analysis NEW!!!; 12. SEVEN SAMURAI OF 2012 (BEST DISCUSSIONS) NEW!!!; 12. Tricky questions from previous years. NEW!!!;

COLLECTION OF QUESTIONS:
PS: 1. Tough and Tricky questions; 2. Hard questions; 3. Hard questions part 2; 4. Standard deviation; 5. Tough Problem Solving Questions With Solutions; 6. Probability and Combinations Questions With Solutions; 7 Tough and tricky exponents and roots questions; 8 12 Easy Pieces (or not?); 9 Bakers' Dozen; 10 Algebra set. ,11 Mixed Questions, 12 Fresh Meat

DS: 1. DS tough questions; 2. DS tough questions part 2; 3. DS tough questions part 3; 4. DS Standard deviation; 5. Inequalities; 6. 700+ GMAT Data Sufficiency Questions With Explanations; 7 Tough and tricky exponents and roots questions; 8 The Discreet Charm of the DS ; 9 Devil's Dozen!!!; 10 Number Properties set., 11 New DS set.


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Re: GMAT Combinatorics 4 [#permalink] New post 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!.

Does this make sense?

Theory on Combinations: math-combinatorics-87345.html

DS questions on Combinations: search.php?search_id=tag&tag_id=31
PS questions on Combinations: search.php?search_id=tag&tag_id=52

Tough and tricky questions on Combinations: hardest-area-questions-probability-and-combinations-101361.html

Theory on probability problems: math-probability-87244.html

All DS probability problems to practice: search.php?search_id=tag&tag_id=33
All PS probability problems to practice: search.php?search_id=tag&tag_id=54
[/quote]


Thanks Bunuel , been doing combinations and probability since morning. Finally getting a hang of it :) . Thanks for the links
Re: GMAT Combinatorics 4   [#permalink] 08 May 2014, 10:28
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