Answer by Austin Mohr for Non-enumerative proof that there are many...
L. Lu and L. A. Szekely have successfully applied the Lopsided (i.e. Negative Dependency Graph) Lovasz Local Lemma to this problem.A negative dependency graph is as a dependency graph except that...
View ArticleAnswer by Brendan McKay for Non-enumerative proof that there are many...
The mean number of fixed points is 1. This is very elementary.Consider the operation of rotating three values around: $p(i)\to p(j)\to p(k)\to p(i)$. Given a permutation with no fixed points, there are...
View ArticleAnswer by Chris Godsil for Non-enumerative proof that there are many...
The number of derangements of an $n$-set is the number of perfect matchingsin the bipartite complement of $n$ disjoint copies of $K_2$. The rook polynomialof $nK_2$ is $(x-1)^2$ and so the number of...
View ArticleAnswer by Robert Israel for Non-enumerative proof that there are many...
The fraction $f(n)$ of permutations of $n$ letters that are derangements satisfies the recurrence $f(n) - f(n-1) = \frac{-1}{n} (f(n-1) - f(n-2))$ with $f(0)=1$ and $f(1)=0$.It is easy to see from this...
View ArticleAnswer by Liviu Nicolaescu for Non-enumerative proof that there are many...
There is a very elegant approach to derangementsD.M. Jackson, Laguerre polynomials and derangements, Math. Proc. Camb. Phil. Soc., 80 (1976), 213–214.I am not sure it fits precisely your criteria, but...
View ArticleAnswer by Barry Cipra for Non-enumerative proof that there are many...
I'm not sure this qualifies, but here goes.It doesn't take any fancy enumerations to show that the average number of fixed points is 1: Among the $n!$ permutations of $n$ objects, each object is fixed...
View ArticleNon-enumerative proof that there are many derangements?
Recall that a derangement is a permutation $\pi: \{1,\ldots,n\} \to \{1,\ldots,n\}$ with no fixed points: $\pi(j) \neq j$ for all $j$. A classical application of the inclusion-exclusion principle tells...
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