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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]

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24 Mar 2007, 16:48
This topic is locked. If you want to discuss this question please re-post it in the respective forum.

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
VP
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24 Mar 2007, 17:47
shoule be (5-1)! = 24
Senior Manager
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24 Mar 2007, 18:45
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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24 Mar 2007, 18:51
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....
Director
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24 Mar 2007, 21:35
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 ..
Manager
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25 Mar 2007, 15:49
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.
Manager
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25 Mar 2007, 17:45
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.

Regards,

Bhaarat
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Re: Permutation/Combination on a Circle [#permalink]

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25 Mar 2007, 21:02
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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27 Mar 2007, 18:23
nitinneha, could you please give the OA?

Thanks!
27 Mar 2007, 18:23
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