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# At a dinner party , 5 people are to be seated around a

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Intern
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At a dinner party , 5 people are to be seated around a [#permalink]

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08 Jan 2008, 06:53
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At a dinner party , 5 people are to be seated around a circular table . Two seating arrangments are considered different only when the positions of the people are different relative to each other. What is the total number of different possible seating arrangments for the group?
A. 5
B. 10
C. 24
D. 32
E. 120

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Director
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08 Jan 2008, 07:00

5! = 120, but since it's a circular table we need to fix one person (5-1)! = 4! = 24. Same technique applies for all permutations where one person needs to be fixed.

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Intern
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09 Jan 2008, 02:02
eschn3am wrote:

5! = 120, but since it's a circular table we need to fix one person (5-1)! = 4! = 24. Same technique applies for all permutations where one person needs to be fixed.

Also can you please advise to me the level of this sort of problem in GMAT ?

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Director
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09 Jan 2008, 02:06
Richard Lee wrote:
At a dinner party , 5 people are to be seated around a circular table . Two seating arrangments are considered different only when the positions of the people are different relative to each other. What is the total number of different possible seating arrangments for the group?
A. 5
B. 10
C. 24
D. 32
E. 120

C.

For circular arrangement, consider one place fixed and find the arrangement for the remaining seats, in this case 4! = 24

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Director
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09 Jan 2008, 04:35
Richard Lee wrote:
eschn3am wrote:

5! = 120, but since it's a circular table we need to fix one person (5-1)! = 4! = 24. Same technique applies for all permutations where one person needs to be fixed.

Also can you please advise to me the level of this sort of problem in GMAT ?

I haven't taken the test before, so I really don't know. I believe that permutation and combination problems are pretty high level though

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CEO
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09 Jan 2008, 20:52
Richard Lee wrote:
At a dinner party , 5 people are to be seated around a circular table . Two seating arrangments are considered different only when the positions of the people are different relative to each other. What is the total number of different possible seating arrangments for the group?
A. 5
B. 10
C. 24
D. 32
E. 120

24. We have to fix one person so its actually 4! I'm not really sure why we have to fix one person. I just know we do. Some more insight on the reasoning behind that would be awsome.

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Manager
Joined: 22 Oct 2007
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13 Jan 2008, 14:20
GMATBLACKBELT wrote:
Richard Lee wrote:
At a dinner party , 5 people are to be seated around a circular table . Two seating arrangments are considered different only when the positions of the people are different relative to each other. What is the total number of different possible seating arrangments for the group?
A. 5
B. 10
C. 24
D. 32
E. 120

24. We have to fix one person so its actually 4! I'm not really sure why we have to fix one person. I just know we do. Some more insight on the reasoning behind that would be awsome.

Consider 5 points chairs on a straight line (A..E) each with one person seated on it. Push each person to the seat next to him. The last one (E) will move to seat A which changes the arrangement (as the relative position of E changes).

Now consider the same 5 chairs and people on a circle. In this case E already has A as his neighbor. moving a person to the chair next to him does not change the arrangement. E still has A as his neighbor and there is no change in the relative position of A and E. Same is the case with rest of the people. A (and similarly B,C,D and E) can occupy one of the 5 chairs without changing their relative position and hence the arrangement for any particular seating sequence. Therefore we need to divide 5! by 5 to arrive at the final answer.
Hope this helps and apologies if it sounds convoluted.

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Manager
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13 Jan 2008, 19:10
4! is wonderful since we can just use it as a formula. dont waste time thinking.

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Manager
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08 Mar 2008, 22:19
Therefore we need to divide 5! by 5 to arrive at the final answer.

I am still having troubles understanding this. Initially When I saw the problem I thought it should be 5!-5. where am i wrong? Thanks!

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CEO
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08 Mar 2008, 22:58
1
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CaspAreaGuy wrote:
Therefore we need to divide 5! by 5 to arrive at the final answer.

I am still having troubles understanding this. Initially When I saw the problem I thought it should be 5!-5. where am i wrong? Thanks!

http://mathworld.wolfram.com/CircularPermutation.html
_________________

You tried your best and you failed miserably. The lesson is 'never try'. -Homer Simpson

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Re: seating problem   [#permalink] 08 Mar 2008, 22:58
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