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Difficult Math Problem #69 - Permutations

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Re: Difficult Math Problem #69 - Permutations

by kulksnikhil » Sat Dec 02, 2006 3:29 am
800guy wrote:In how many ways can 5 people sit around a circular table if one should not have the same neighbors in any two arrangements?
5P5 = 5
Answer is 5
Last edited by kulksnikhil on Sat Dec 02, 2006 3:34 am, edited 1 time in total.
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by lalitaroral » Sat Dec 02, 2006 3:32 am
If no condition applied, Total ways= 5!= 120

After condition

Lets calculate the ways in which same neighbour can occur

Lets fix one person, he can be srrounded by 2 ppl and rest 2 can be arranged in 2! way

probability of choosing left n right side person = 4*3

Total possible ways= 2!*4*3=24

So, number of wayz the person will not be same = 120-24=96
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OA

by 800guy » Mon Dec 04, 2006 2:42 pm
OA

The ways of arranging 5 people in a circle = (5-1)! = 4!
For a person seated with 2 neighbors, the number of ways of that happening is 2:
AXB or BXA, where X is the person in question.
So, for each person, we have two such arrangements in 4!. Since we don't want to repeat arrangement, we divide 4!/2 to get 12
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by r321 » Sun Jan 02, 2011 12:34 pm
800guy wrote:In how many ways can 5 people sit around a circular table if one should not have the same neighbors in any two arrangements?
I only see two possible arrangement: ABCDE ('A' has neighbours 'B' & 'E') & ACEBD ('A' has neighbours 'C' & 'D').

Could some one please list out any of the other 10 arrangements because I don't see how they can be rearranged more than once without repeating neighbours.
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by Night reader » Sun Jan 02, 2011 3:14 pm
r321 wrote:
800guy wrote:In how many ways can 5 people sit around a circular table if one should not have the same neighbors in any two arrangements?
I only see two possible arrangement: ABCDE ('A' has neighbours 'B' & 'E') & ACEBD ('A' has neighbours 'C' & 'D').

Could some one please list out any of the other 10 arrangements because I don't see how they can be rearranged more than once without repeating neighbours.
@r321, AXB and BXA are considered two different arrangements with two different neighbors. According to this logic, the neighbors on the left and right sides are shifting (permuting); we have A-B-C-D-E and A-C-E-B-D as well as C-B-A-D-E, C-B-D-E-A ...

if we continue our selection from the possible (5-1)! circular combination set, we get exactly 12.
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by r321 » Sun Jan 02, 2011 9:35 pm
Night reader wrote:
r321 wrote:
800guy wrote:In how many ways can 5 people sit around a circular table if one should not have the same neighbors in any two arrangements?
I only see two possible arrangement: ABCDE ('A' has neighbours 'B' & 'E') & ACEBD ('A' has neighbours 'C' & 'D').

Could some one please list out any of the other 10 arrangements because I don't see how they can be rearranged more than once without repeating neighbours.
@r321, AXB and BXA are considered two different arrangements with two different neighbors. According to this logic, the neighbors on the left and right sides are shifting (permuting); we have A-B-C-D-E and A-C-E-B-D as well as C-B-A-D-E, C-B-D-E-A ...

if we continue our selection from the possible (5-1)! circular combination set, we get exactly 12.
But the question clearly says "one should not have the same neighbours". If B has been seated adjacent to X (BXA) in one arrangement it can't be seated next to X again regardless of whether it's on the left or right of X, because being seated adjacent to X makes it a neighbour of X again.

BXA & AXB could be considered different if the question said something like "one should not have the same neighbours to the left/right"

Also, in A-B-C-D-E and C-B-A-D-E, D & E are being seated next to each other in the same clockwise order twice, which is not allowed according to my understanding of the restriction. The same applies for A-C-E-B-D and C-B-D-E-A, in which B & D are neighbours in the same clockwise order twice.

Maybe I'm failing to understand the logic here. If I am wrong please help me understand how by listing out the arrangements because I still don't see more than two possible arrangements considering the restrictions. I also do understand that these questions from the 'gmat MATH tough problems' doc are worded very poorly, which would never be the case with an official GMAT problem. Perhaps, the confusion over the different solutions to this problem is due to differing interpretations of the question.
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by Night reader » Sun Jan 02, 2011 10:01 pm
r321 to state data and seek solution we need one- may be two words; enough words are embedded into 198+ quest level 700. Please remember one restriction for this problem, namely circular permutation. You move either clock-wise or counter-clockwise along the circle. Thus you should seek: one neighbor following OR one preceding but not one neighbor following AND one preceding. If I'm wrong let some one correct me.
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by RACHVIK » Sun Jan 02, 2011 11:51 pm
Can some expert advise?? I feel the total arrangements should be 2.
Rachvik
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by Night reader » Sun Jan 02, 2011 11:57 pm
RACHVIK wrote:Can some expert advise?? I feel the total arrangements should be 2.
you feel or calculate? ;) both directions you are bet, one direction it's 12
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