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

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5 noble knights are to be seated at a round table. In how [#permalink] New post 07 May 2008, 13:02
5 noble knights are to be seated at a round table. In how many ways can they be seated?

* 120
* 96
* 60
* 35
* 24


Since there are no arrangement issues, shouldn't that be a 5 * 5 * 5 * 5 * 5 deal?
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Re: Prob - [#permalink] New post 07 May 2008, 13:05
I have not studied prob yet. I believe it should be 5! or 5*4*3*2*1=120. A You repeat combinations in your scenario. Try it with two knights and seats. The answer is not 4.
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Re: Prob - [#permalink] New post 07 May 2008, 13:13
Maple wrote:
I have not studied prob yet. I believe it should be 5! or 5*4*3*2*1=120. A You repeat combinations in your scenario. Try it with two knights and seats. The answer is not 4.




circular probability = (n-1)! , so 4! = 24
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Re: Prob - [#permalink] New post 07 May 2008, 13:24
I would have never known. Thanks
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Re: Prob - [#permalink] New post 07 May 2008, 13:31
In the two knight scenario, wouldn't the knights be able to trade seats for a total of two combinations instead of the one seating arrangement that the (n-1)! formula would suggest.
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Re: Prob - [#permalink] New post 07 May 2008, 14:16
Maple wrote:
In the two knight scenario, wouldn't the knights be able to trade seats for a total of two combinations instead of the one seating arrangement that the (n-1)! formula would suggest.



I see your point , but then I think can you have a circular arrangement with just two combination , I mean if you think about it mathematically two arrangemnts will always form a straight line , so no cicular arrangement here . does it make sense?
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Re: Prob - [#permalink] New post 07 May 2008, 15:26
rpmodi wrote:
I see your point , but then I think can you have a circular arrangement with just two combination , I mean if you think about it mathematically two arrangements will always form a straight line , so no cicular arrangement here . does it make sense?


Depends on whether the actual seat the knight sits in matters or whether the question is more interested in the physical order of the knights. right?
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Re: Prob - [#permalink] New post 07 May 2008, 16:04
Maple wrote:
rpmodi wrote:
I see your point , but then I think can you have a circular arrangement with just two combination , I mean if you think about it mathematically two arrangements will always form a straight line , so no cicular arrangement here . does it make sense?


Depends on whether the actual seat the knight sits in matters or whether the question is more interested in the physical order of the knights. right?


May be this will clear your doubt .

http://www.tutorvista.com/content/math/ ... ations.php

if you take example of 2 knights O1O2 arrangement will be same as O2O1 arrangement , so the question is asking for position of knights relative to other knights in the group .
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Re: Prob - [#permalink] New post 08 May 2008, 09:40
so is the OA 120 using 5!?
Re: Prob -   [#permalink] 08 May 2008, 09:40
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