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In how many ways 30 couples(total=30 men and 30 women...with

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In how many ways 30 couples(total=30 men and 30 women...with [#permalink] New post 19 Aug 2003, 12:29
In how many ways 30 couples(total=30 men and 30 women...with respective man sitting on the right of woman) and 30 HOD(head of the department) be seated around a round table if each of the couple should sit between each of the HOD?
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 [#permalink] New post 19 Aug 2003, 13:22
heres my try at this kind of problem:
we fix one HOD... the others can distribute in 29! ways... now we have to distribute the couples... i understand that the couples are fixed (ie to each woman theres a man assigned)... so they can arrange in 30! ways (as men have to sit on the right)... so anwer would be 29!*30!?

now, how would it be if the couples were not fixed (any man can sit behind any woman)?
we can make 15! couples, so answer would be 29!*15!*30!?

and finally, what if the positions of the couples were not fixed (ie man can sit either on the right or on the left)...
answer would be 29!*15!*30!*2^30?

s'one brighter than me please check my numbers... cheers, javi
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Re: Probability Q3 [#permalink] New post 22 Aug 2003, 22:47
aps_can wrote:
In how many ways 30 couples(total=30 men and 30 women...with respective man sitting on the right of woman) and 30 HOD(head of the department) be seated around a round table if each of the couple should sit between each of the HOD?


Agree with 29!*30!
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 [#permalink] New post 25 Aug 2003, 03:33
javropu wrote:
heres my try at this kind of problem:
we fix one HOD... the others can distribute in 29! ways... now we have to distribute the couples... i understand that the couples are fixed (ie to each woman theres a man assigned)... so they can arrange in 30! ways (as men have to sit on the right)... so anwer would be 29!*30!?

now, how would it be if the couples were not fixed (any man can sit behind any woman)?
we can make 15! couples, so answer would be 29!*15!*30!?

and finally, what if the positions of the couples were not fixed (ie man can sit either on the right or on the left)...
answer would be 29!*15!*30!*2^30?

s'one brighter than me please check my numbers... cheers, javi


If any man can sit next to any woman, assuming the men still sit to the same side next to each woman, the answer would be simply:

29! x 30! x 30!
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Vice President, Midtown NYC Investment Bank, Structured Finance IT
MFE, Haas School of Business, UC Berkeley, Class of 2005
MBA, Anderson School of Management, UCLA, Class of 1993

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 [#permalink] New post 25 Aug 2003, 03:45
thx akamai :) i messed it all up... for calculating the number of couples, i took 15 instead of 30... oh my... its interesting to see that you can get to the same result by i) calculating the number of couples (30!) and then calculating how to arrange the HOD and the couples (29!*30!) or ii) by just arranging the 3 people (29!*30!*30!)
  [#permalink] 25 Aug 2003, 03:45
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