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July 2nd, 2017, 07:57 PM  #1 
Senior Member Joined: Jul 2011 Posts: 395 Thanks: 15  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: 17,533 Thanks: 1322 
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? 
July 3rd, 2017, 07:49 AM  #3 
Math Team Joined: Jul 2013 From: काठमाडौं, नेपाल Posts: 464 Thanks: 37 Math Focus: Elementary Mathematics 
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: 395 Thanks: 15 
To moderator ACEBD and CEBDA counted as 1, Thanks

July 25th, 2017, 05:28 AM  #5 
Math Team Joined: Jan 2015 From: Alabama Posts: 2,524 Thanks: 643 
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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