# Why is it trivial to see that $\binom{n}{2} = \frac{n!}{(n-2m)!m!2^m}$ has no solutions for $m > 3$?

I am currently working through an exercise in Dummit & Foote, in which I try to prove that every automorphism of $S_n$ with $n \geq 3, n \neq 6$ is an inner automorphism ($S_n$ being the permutation group of the set $\{1,\ldots, n\}$). I showed that the number of permutations equals $n(n-1)/2$ and the number of elements of order $2$ which are the product of $m$ disjoint transpositions equals $$\frac{n!}{(n - 2m)!m!2^m}.$$ I tried to find solutions for $m$, but did not succeed and then stumbled on this question. I understand all of the accepted answer, but the last step: is it trivial to see that there could never be any solutions for $m > 3$?

any help would be appreciated.

• Your equation can be written as $\binom{n}{2}=\binom{n}{2m}\cdot\frac{(2m)!}{m!\cdot2^m}$. It can be shown by induction that for $m>3$, RHS > LHS. – TheSimpliFire Oct 23 '17 at 10:42

In the linked question, the claim is that $\frac{m}{(2m-3)!!}$ is never an integer for $m>3$. This is true because $2m-3>m$, so $$0<\frac{m}{(2m-3)!!}\leq \frac{m}{2m-3}<1,$$ and of course there can be no integer between $0$ and $1$...
• I am just assuming this; is $n!!$ equal to $(n!)!$ or is it something different? – Mr Pie Oct 23 '17 at 10:47
• @user477343 No, here $n!!$ is the double factorial : it is defined by taking the product of all natural less than or equal to $n$ that have the same parity as $n$. So $(2m-3)!!=1\cdot 3\cdots (2m-5)\cdot (2m-3)$. See also en.wikipedia.org/wiki/Double_factorial – Arnaud D. Oct 23 '17 at 10:58