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How can I show that the amount of possible products of disjoint transpositions in $S_n$ is strictly more than two times the amount of possible products of even number of disjoint transpositions in $A_n$ for every $n>3$?

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The possible products of disjoint transpositions straightly correspond to collection of disjoint couples of $\{1,2,...,n\}$. To form $k$ couples ($k\le n/2$) can be done in $$ T_{n,k}:= \frac{\overbrace{\binom n2 \binom{n-2}2 \ldots \binom{n-2k+2}2}^k}{k!}$$ ways. This can also be written as $$T_{n,k} = \binom n{2k}\cdot\frac{(2k)!}{k!}\cdot \frac1{2^k} = \binom n{2k}\cdot (2k-1)(2k-3)\ldots\cdot3\cdot 1$$ Probably this helps.

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