7
votes
1answer
292 views

Non-enumerative proof that there are many simple permutations?

Terence Tao asked for a non-enumerative proof that a positive proportion of permutations are derangements and got a great answer. Inspired by this, I'd like to ask about another family of ...
6
votes
0answers
195 views

Counting Selections of Entries such having an Extremal Permutation of length n^2+1

Let $S_{n^2+1}$ be permutations of length $n^2+1$. By Erdos-Szekeres Theorem. any $s \in S_{n^2+1}$ would have a monotone subsequence(increasing or decreasing)of length $n+1$. Say a permutation $s$ of ...
36
votes
6answers
3k views

Non-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 ...