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5 noble knights are to be seated around a round table. In

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Manager
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5 noble knights are to be seated around a round table. In [#permalink] New post 24 Mar 2007, 15:48
5 noble knights are to be seated around a round table. In how many ways can they be seated?

1) 120
2) 96
3) 60
4) 35
5) 24
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 [#permalink] New post 24 Mar 2007, 16:47
shoule be (5-1)! = 24
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 [#permalink] New post 24 Mar 2007, 17:45
Summer3 wrote:
Why not 5! = 120???


Hmmm...

I think it's because they're around a "round" table. :?

So have one knight "fixed" and find the combination for the other four...

4! = 24

Since it's round, the fixed person will still have the effect of being "moved around" because those on either sides of him will continuously move.

5! would be the case for a row, where there is no one to the left and right of those at the ends...
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 [#permalink] New post 24 Mar 2007, 17:51
Yeah, I also think so but still need some clarification and confirmation :)

This is the first time I saw such question....combinations but not in a linear way....
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 [#permalink] New post 24 Mar 2007, 20:35
hi, didnt the q say if clockwise and anticlockwise are to be treated as same or different ?

the answer will vary depending on the above condition ..
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 [#permalink] New post 25 Mar 2007, 14:49
grad_mba wrote:
hi, didnt the q say if clockwise and anticlockwise are to be treated as same or different ?

the answer will vary depending on the above condition ..


same thoughts .. not enough data ...

but so far (A) 5! = 120. it seems to be the same case such as staying in a line = permutation with order.
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Permutation/Combination on a Circle [#permalink] New post 25 Mar 2007, 16:45
When arrangements are to be made in a circle and we have n objects,we use (n-1)! factorial. This is becouse we have to have a reference point when making arrangements.
The same rule applies to permutation problems.

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Re: Permutation/Combination on a Circle [#permalink] New post 25 Mar 2007, 20:02
bhaarat wrote:
When arrangements are to be made in a circle and we have n objects,we use (n-1)! factorial. This is becouse we have to have a reference point when making arrangements.
The same rule applies to permutation problems.

Regards,

Bhaarat


still do not understand why should we have the "reference point"?
i believe the ref. point could have any 'knight'. i mean we fix the 'start ref. point' but the place can be updated by any knight.

so 5 slots in a circle, then 5! permutations.
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 [#permalink] New post 27 Mar 2007, 17:23
nitinneha, could you please give the OA?

Thanks!
  [#permalink] 27 Mar 2007, 17:23
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