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 Post subject: Probabilty - At a dinner party, 5 people are to be seated...
 Post Posted: Mon May 26, 2008 4:24 pm 
At a dinner party, 5 people are to be seated around a circular table. Two seating arrangements are considered different only when the positions of people are different relative to each other. What is the total no. of different possible seating arrangements for the group?
Ans: 24

I think I need to really polish my propbability skills... I really need help on this one. I guess, I am not getting the concept right. Can you please suggest me how this should be solved?

Cheers
~a


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 Post subject: Re: Probabilty - At a dinner party, 5 people are to be seate
 Post Posted: Wed May 28, 2008 7:41 am 
ameya wrote:
At a dinner party, 5 people are to be seated around a circular table. Two seating arrangements are considered different only when the positions of people are different relative to each other. What is the total no. of different possible seating arrangements for the group?
Ans: 24

I think I need to really polish my propbability skills... I really need help on this one. I guess, I am not getting the concept right. Can you please suggest me how this should be solved?

Cheers
~a

fix one and rest can be arranged in 4! ways that is 24


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 Post subject: Re: Probabilty - At a dinner party, 5 people are to be seate
 Post Posted: Wed Jun 04, 2008 7:31 am 
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ManhattanGMAT Staff


Posts: 13427
sanj wrote:
fix one and rest can be arranged in 4! ways that is 24


well put.

the difference between a round table and a normal table is that a round table has no ends. therefore, you can arbitrarily select which seat is to be called seat #1.
two ways to approach this problem:

(1)
as the poster above has said, you can fix one of the people in place (or, equivalently, rotate the table like a carousel so that that person always winds up in the same place). this will avoid the production of multiple equivalent scenarios in which the people are seated in the same order, but have just shuffled a seat or two over (these don't count as different arrangements).
then you have free rein to arrange the other four people, so that's 4! = 24 arrangements.

(2)
alternatively, you can figure out that there are 5! = 120 arrangements overall.
however, each UNIQUE arrangement is actually repeated five times: because there are five seats at the table, there are 5 different versions of every possible seating arrangement. (for instance, ABCDE, BCDEA, CDEAB, DEABC, EABCD are all equivalent.)
so this means you must divide by 5: 120 / 5 = 24


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 Post subject: Re: Probabilty - At a dinner party, 5 people are to be seated...
 Post Posted: Mon Nov 07, 2011 3:06 pm 
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Forum Guests


Posts: 1
I have been reviewing this problem, and I think maybe I don't understand understand the meaning of "different relative to each other." Can you explain to me why each unique arrangement is repeated five times, and not ten? For the example Ron gave:

ABCDE, BCDEA, CDEAB, DEABC, EABCD but not
EDCBA, AEDCB, BAEDC, CBAED, DCBAE

To me, placing people in the same order, whether clockwise or counterclockwise, seems to be the same position relative to each other. Everyone is sitting next to the same two people.

My answer to this question, on the test and in review, is perpetually "12." What am I missing here?


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 Post subject: Re: Probabilty - At a dinner party, 5 people are to be seated...
 Post Posted: Wed Nov 09, 2011 7:00 am 
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ManhattanGMAT Staff


Posts: 13427
npowell43 wrote:
Quote:
To me, placing people in the same order, whether clockwise or counterclockwise, seems to be the same position relative to each other. Everyone is sitting next to the same two people.


if you are to my left and my wife is to my right (first five arrangements), that's a different arrangement than if you are to my right and she is to my left (last five).

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 Post subject: Re: Probabilty - At a dinner party, 5 people are to be seated...
 Post Posted: Tue Jul 31, 2012 2:44 pm 
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Students


Posts: 16
I would take this opportunity to explain the concept, which Ron has already stated above, in little detail.

This concept is called circular permutation. If we imagine for e.g. 4 persons sitting around a table, then for each of their circular arrangement we can get 4 linear arrangements.
If the 4 persons are labeled as 1,2,3,4 sitting in clock wise direction around the table, then the 4 linear arrangement would be 1234,2341, 3412, and 4123.

If we change the sitting arrangement of the 4 persons around the table as 3214 in clock wise direction, then the linear arrangements would be

3214, 2143, 1432, and 4312.
Thus we find that for each circular arrangement we get 4 linear arrangements. Let the total number of circular arrangements be c. for each circular arrangements we get n linear arrangements for n persons sitting around the table. Therefore the total linear arrangement would be n*c. However we already know that the exhaustive number of cases of arranging the people linearly is n!.
=> n!= n*c
so c = n!/n = (n-1)!
So total circular permutations is (n-1)! where the clockwise and counter clockwise directions are considered different arrangements. If the clockwise and counter clockwise arrangements are not different, then the total no. of circular permutations become (n-1)! /2.

Here there are 5 people sitting around the table. Since counter clockwise and clockwise directions are considered different, so the different circular permutations will be (5-1)! = 24.

Thanks
Prashant
Big Fan!!


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 Post subject: Re: Probabilty - At a dinner party, 5 people are to be seated...
 Post Posted: Thu Aug 02, 2012 7:31 am 
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ManhattanGMAT Staff


Posts: 13427
yeah.

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