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gmatfordays
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Praetorian
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Paul
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Paul
I'd say (10-1)! / 2 because you can sit them clockwise or counterclockwise which would be the same



the way , you solve these kind of problems is to fix one person and then calculate the number of ways the others can be arranged

with 2 persons, fix one, you will have 1! ways

with 4 , fix one, you will have 3! ways

similarly with n, fix one, you will have (n-1)! ways

thanks
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Paul
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Hey Praethorian, (10-1)! is what I originally thought but here is a link to the answer I told you.
https://www.testmagic.com/forum/topic.asp?TOPIC_ID=1699
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Solo
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Paul, good job with reference, but I pick up 9! as correct.

From your standpoint, you know, anti-clockwise and clockwise arrangements are NOT the same in this particular case.

Kpadma, I am in the same boat with 9!
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Paul
Hey Praethorian, (10-1)! is what I originally thought but here is a link to the answer I told you.
https://www.testmagic.com/forum/topic.asp?TOPIC_ID=1699


The clockwise anticlockwise symmetry would hold true from something that you could "flipover" (e.g., charm bracelet, hula hoop, plain rings, etc) and not be able to tell which side is up.
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Nice example. Thx Akamai
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AkamaiBrah
Paul
Hey Praethorian, (10-1)! is what I originally thought but here is a link to the answer I told you.
https://www.testmagic.com/forum/topic.asp?TOPIC_ID=1699

The clockwise anticlockwise symmetry would hold true from something that you could "flipover" (e.g., charm bracelet, hula hoop, plain rings, etc) and not be able to tell which side is up.


Yes, very good explanation.. If you look at the link that Paul provided you find that the original poster made a mistake, so hopefully what I wrote at the end makes it more clear. This made me fumble, hence why I posted it to this board.



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