I calculated the number of permutations with no 2-cycles in two ways but I got 2 different results I calculated the number of permutations in $S_n$ with no 2-cycles in two ways but I got 2 different results. The first time I used the principle of inclusion-exclusion and I got $\sum_{k=0}^n \frac{n!}{k!}\frac{1}{2^k}(-1)^k$ and I'm pretty sure that it's right. The second way is using generating functions. Using the exponential formula I calculated that the generating function of this type of permutations is $\frac{e^{-x^2/2}}{1-x}$. So it is $(\sum x^n)(\sum 1/n! (-1/2)^n x^{2n}$. If I make the product of these two series I got the series with coefficient $\sum_{k=0, k\;even}^n 1/(k/2)!(-1/2)^{k/2}$. Could you tell me where is the mistake?
This is the computation of inclusion-exclusion: 
we count the permutations with at least one 2-cycle. The number of permutation with at least k 2-cycles is $c_k=(n-2k)!\frac{\binom{n}{2}\binom{n-2}{2}\cdots\binom{n-2k+2}{2}}{k!}$. So the permutations with at least one 2-cycle are $\sum_{k=1}^n(-1)^kc_k$. So what we want is $n!-\sum_{k=1}^n\frac{n!}{k!}\frac{1}{2^k}(-1)^k=\sum_{k=0}^n\frac{n!}{k!}(-1/2)^k$. So it's probable that the mistake is in the computation of $(-1)^kc_k$. Does anyone see an expression for this?

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