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Let $[n]$‎ ‌‎be a set with $n$ element‎. ‎A permutation $\sigma$ of the symmetric group $S_n$ is called a derangement of $[n]$ if $\sigma(i)\neq i$ for each $i\in[n]$‎. ‎Suppose that $\theta$ and $\gamma$ are two arbitrary permutations of $S_n$‎. ‎We say that a permutation $\sigma$ is a double derangement with respect to $\theta$ and $\gamma$ if $\sigma(i)\neq \gamma(i)$ and $\sigma(i)\neq \theta(i)$ for each $i\in[n]$ and denoted by $D_n(\gamma,\theta)$. Could you please give explicit formula or recurrence relation for the number of such derangement ($|D_n(\gamma,\theta)|$?

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d.y
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  • I assume you mean $D_n(\gamma,\theta)$, rather than $D_n(\sigma,\theta)$? The variable $\sigma$ is unbound, so it doesn't make much sense in that context. Also, when you say 'a formula for $D_n(\gamma,\theta)$', do you really mean a formula for $\lvert D_n(\gamma,\theta)\rvert$? – Nick Peterson Feb 07 '18 at 17:17
  • Right, I mean $D_n(\gamma, \theta)$. The formula means recurrence relation or the number of such derangement ($|D_n(\gamma,\theta)|$) – d.y Feb 07 '18 at 17:27
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    This possibly depends on $\gamma$ and $\theta$. It's basically the problem of counting how many possible third rows of a Latin square there are, given the first two rows. – Angina Seng Feb 07 '18 at 17:58
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    Yeah, it will definitely depend on (at minimum) the number of $i$ such that $\gamma(i)=\theta(i)$. – Nick Peterson Feb 07 '18 at 18:01
  • @LordSharktheUnknown How the number of such derangement is corresponding of such Latin square? could you please give example? – d.y Feb 07 '18 at 18:03
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    By computer counting, the number of double derangements with $n=5$ is either $12,13,16,19,24$ or $44$. The probability of these numbers is $20/120,24/120,45/120,20/120,10/120$ and $1/120$ – Empy2 Feb 08 '18 at 04:12
  • On average, it seems to be $n!/e^2$, but if $\gamma=\theta$, it can be as much as $n!/e$. – Empy2 Feb 08 '18 at 04:39
  • @LordSharktheUnknown How can obtain recurrence relation or explicit formula for this derangment? – d.y Feb 08 '18 at 17:23
  • Already asked here (can't be closed as a duplicate because there's no upvoted or accepted answer). – joriki Dec 21 '19 at 20:57

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