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 July 2nd, 2017, 07:57 PM #1 Senior Member   Joined: Jul 2011 Posts: 405 Thanks: 16 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.
 July 2nd, 2017, 11:00 PM #2 Global Moderator   Joined: Dec 2006 Posts: 20,978 Thanks: 2229 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? Thanks from panky
 July 3rd, 2017, 07:49 AM #3 Math Team     Joined: Jul 2013 From: काठमाडौं, नेपाल Posts: 901 Thanks: 61 Math Focus: सामान्य गणित 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.
 July 13th, 2017, 07:16 PM #4 Senior Member   Joined: Jul 2011 Posts: 405 Thanks: 16 To moderator ACEBD and CEBDA counted as 1, Thanks
 July 25th, 2017, 05:28 AM #5 Math Team   Joined: Jan 2015 From: Alabama Posts: 3,264 Thanks: 902 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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