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July 2nd, 2017, 07:57 PM   #1
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counting problem

If $n$ persons are sitting around a table $(n\geq 4)$, then the number of arrangements in which all shall not have same neighbours is

Last edited by skipjack; July 2nd, 2017 at 10:37 PM.
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July 2nd, 2017, 11:00 PM   #2
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If $n$ is 5 and the original arrangement is denoted by ABCDE, are the new arrangements ACEBD and CEBDA counted as just one rearrangement?

In a new arrangement, is it okay that just one neighbour is different, rather than both?
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July 3rd, 2017, 07:49 AM   #3
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Does it mean "everyone has new neighbors" or "at least one(two) has different neighbor"?

Last edited by skipjack; July 3rd, 2017 at 08:21 AM.
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July 13th, 2017, 07:16 PM   #4
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To moderator ACEBD and CEBDA counted as 1, Thanks
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July 25th, 2017, 05:28 AM   #5
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That doesn't answer the question. Do you want the number of orders in which some of the people do not have the same neighbor, or in which none of the people have the same neighbor?
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