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The question goes like: There are n people in a row, how many ways are there to rearrange them, such that for all $0 \le i \le n-1$, the people who stood in original (i+1)th place, is not standing right next to the original people that was in i'th place.

My approach: i was thinking about inclusion exclusion principle. |Ai| - the number of ways to rearrangee the people in a row, such that i and i+1 standing together. now im stuck on the case of $|A_i \cap A_j|$, because there are two possibilities, for example, if i have the row $1 2 3456$, i can either choose 12 34, or 12 23. How can i overcome this issue? thanks in advance, because I'm not certain on how to calculate $|A_i \cap A_j|$ and so on.

csmathstudent8
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  • What does "right to" mean? Do you mean immediately next to and to the right or do you simply mean somewhere to the right? In any case, a natural start is to solve the problem for several small $n$. If nothing else, that gives you a way to check your eventual formula. – lulu Jun 14 '24 at 18:17
  • @lulu it next to, ill fix it – csmathstudent8 Jun 14 '24 at 18:20
  • So, then, this is a duplicate, yes? – lulu Jun 14 '24 at 18:23
  • @lulu i think it's indeed the same, but the answer was not using PIE, im looking into PIE. – csmathstudent8 Jun 14 '24 at 18:29
  • this is A000255. I would think recursive methods are the way to go here. – lulu Jun 14 '24 at 18:47
  • Note: post edit, I still don't understand the rules. Please try to be clear. Writing "right next to", as you do, would exclude both $(i, i+1)$ and $(i+1,i)$. In the first version of your post, you appeared to only want to exclude $(i, i+1)$ and that's what the duplicate I linked to considers. I suggest: Work this out explicitly for $n=3$ and add that to your post, just to remove ambiguities. If you really mean to exclude both orders, then the answer is $0$, right? But if you only mean what is meant in the duplicate then $132, 213, 321$ are all good making the answer $3$. – lulu Jun 14 '24 at 18:50
  • Can you please explain the approach with the recursive method? – csmathstudent8 Jun 14 '24 at 18:54
  • I don't think I could add much to the posted solution. – lulu Jun 14 '24 at 18:55
  • Oh im sorry it's exactly the same, thanks again lulu – csmathstudent8 Jun 14 '24 at 18:55
  • Every problem is not a piece of PIE ! – true blue anil Jun 14 '24 at 19:48
  • @trueblueanil I strongly suspect that this particular problem is solvable (analytically) by Inclusion-Exclusion, through the use of Stars and Bars internally, in the computation of $~T_2, T_3, \cdots.~$ See this quasi-answer. – user2661923 Jun 14 '24 at 21:46
  • @trueblueanil I must admit defeat, and withdraw my previous comment. I tried and failed to use the methods in my previous comment to attack this problem. I started by letting $~S_k~$ denote the set of permutations where a violation occurred at position $~k,~$ where $~k \in {1,2,\cdots,n-1}.~$ The difficulty is that when computing one of the groups of terms in $~T_r,~$ by scrutinizing the variables $~x_2,\cdots,x_r,~$ it is insufficient to determine how many of them are exactly equal to $~0.~$ You must further examine how these zero variables are positioned. ...see next comment – user2661923 Jun 14 '24 at 23:19
  • For example, when computing $~T_4,~$ the enumeration of $~S_1 \cap S_2 \cap S_4 \cap S_5~$ will be different than the enumeration of $~S_1 \cap S_2 \cap S_3 \cap S_5.~$ I also considered redefining $~S_k~$ to denote the permutations where the elements $~k,k+1~$ occurred consecutively, somewhere in the permutation. Unfortunately, I also wasn't able to find a way to successfully use Stars and Bars internally, with this alternate approach. – user2661923 Jun 14 '24 at 23:29
  • @user2661923 what do you mean by $S1∩S2∩S3∩S5.$? – csmathstudent8 Jun 15 '24 at 08:07
  • @csmathstudent8 that is a complicated question. In one of my comments to this posted question, you will find a link to a quasi-answer. Within that link, you will find two Inclusion-Exclusion links. Please examine those two Inclusion-Exclusion links, and then leave me another comment, if you still have a question. – user2661923 Jun 15 '24 at 17:20

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