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I need to find the number of permutations of $n$-items where each of them has cycle of only even length. This is my attempt:

Using the basic combinatorial construction, I have --

\begin{align} \mathcal{\tilde{I}} = \text{SET}(\text{CYC}_{2,4,6,\cdots}(\mathcal{Z})) \end{align} Now using the exponential generating function: \begin{align} \tilde{I}(z) &= \exp\left(\frac{z^2}{2} + \frac{z^4}{4} + \frac{z^6}{6} + \cdots\right)\\ &= \exp\left(\frac{1}{2}\left(\log\left(\frac{1}{1+z}\right) + \log\left(\frac{1}{1-z}\right)\right)\right)\\ &= \left(\frac{1}{1 - z^2}\right)^{\frac{1}{2}} \end{align} I need to show that the coefficient of the GF found from above is $(1^2 \cdot 3^2 \cdot 5^2 \cdots (2n-1)^2)$

I was trying to do it using the Taylor expansion, but things are getting extremely hairy, is there any simple way to do that? I also tried using $x (d/dx) \log$ way, this makes the stuff even more complicated.

Any pointer will be appreciated.

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Hint- We have that for $|z|<1$, $$(1-z^2)^{-\frac{1}{2}}=\sum_{n=0}^{\infty}\binom{-\frac{1}{2}}{n}(-1)^nz^{2n}$$ Now expand, $$\binom{-\frac{1}{2}}{n}(-1)^n=(-1)^n\frac{(-\frac{1}{2})(-\frac{1}{2}-1)\cdots (-\frac{1}{2}-n+1)}{n!}.$$ Note that this is an exponential generating function, hence you are looking for $$(2n)!\binom{-\frac{1}{2}}{n}(-1)^n.$$

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  • $\begingroup$ got it, thanks so much. $\endgroup$
    – ramgorur
    Oct 17, 2016 at 6:12

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