Probabilty of two permutations having common elements? - MathOverflow most recent 30 from http://mathoverflow.net 2013-05-19T07:14:47Z http://mathoverflow.net/feeds/question/97208 http://www.creativecommons.org/licenses/by-nc/2.5/rdf http://mathoverflow.net/questions/97208/probabilty-of-two-permutations-having-common-elements Probabilty of two permutations having common elements? jgonagle 2012-05-17T10:23:15Z 2012-11-04T16:21:49Z <p>What is the probability of two permutations on set X of size m (i.e. |X|=m) having at least n points of intersection? By this I mean that if two permutations, which I'll call g(x) and h(x), map a member x to g(x) for the first permutation and x to h(x) for the second permutation, what is the chance that g(x)=h(x) for at least n values of x?</p> http://mathoverflow.net/questions/97208/probabilty-of-two-permutations-having-common-elements/98606#98606 Answer by Patricia Hersh for Probabilty of two permutations having common elements? Patricia Hersh 2012-06-01T20:40:15Z 2012-11-04T16:21:49Z <p>You can use inclusion-exclusion to show that the number of permutations in $S_m$ having at least $n$ fixed points is $$\sum_{k=n}^m (-1)^{k-n}{k-1\choose n-1}{m\choose k}(m-k)! = m! \sum_{k=n}^m (-1)^{k-n}\frac{1}{k!}{k-1\choose n-1}$$ </p> <p>We describe now how to obtain the lefthand expression. The ${m\choose k}$ comes from choosing $k$ fixed points, the $(m-k)!$ counts permutations in $S_m$ having these $k$ fixed points, and then $(-1)^{k-n}{k-1\choose n-1}$ is an inclusion-exclusion counting coefficient, namely the Möbius function $\mu (\hat{0},\hat{1})$ on the subposet of the Boolean algebra of subsets of ${ 1,\dots ,k }$ where we exclude the subsets having size $1\le i \le n-1$. </p> <p>One way to calculate this Möbius function is to use that each rank-selection of the Boolean algebra is lexicographically shellable. The desired Möbius function will be $(-1)^{k-n}$ multiplied by the number of so-called descending chains'' in the lexicographic shelling, which in this case is the number of permutations in $S_k$ that are ascending in the first $n$ letters and then descending after that, which in particular forces the letter $k$ to be the $n$-th letter in the permutation (in one-line notation). </p> <p>This includes the well-known special case (usually phrased in terms of derangements) that the number of permutations in $S_m$ with at least one fixed point is $\sum_{k = 1}^m (-1)^{k-1} {m\choose k}(m-k)!$ which equals $- (-m! + \sum_{k= 0}^m(-1)^k {m\choose k}(m-k)!)$. As $m$ goes to infinity, this approaches $-m!(-1 + 1/e) = m!(1-1/e)$. A good reference for the $n=1$ case is chapter 2 of Enumerative Combinatorics, Volume 1, by Richard Stanley. The original source for lexicographic shellability is Anders Björner's paper Shellable and Cohen-Macaulay partially ordered sets''.</p> <p><strong>Added later:</strong> the comments above mention the recontres numbers. These give an approach for obtaining the $n>1$ case as a consequence of the $n=1$ case -- by choosing your fixed point set and then counting derangements on the remaining letters, summing over the possible fixed point sets. This results in a double sum, with an alternating sum as the inner sum. </p>