Counting permutations in $S_n$ with $k_i$ cycles of length $i$ using Polya's theorem I had this assignment on counting the number of permutations in $S_n$ with $k_i$ cycles of length $i$. It is pretty easy to find that the answer is $\frac{n!}{1^{k_1}2^{k_2}\cdot\cdot\cdot k_1!k_2!\cdot\cdot\cdot k_n!}$ just by actually counting them. 
I also tried to do this using Polya's theorem, but so far without success. The idea is to reformulate the problem as a coloring problem (and then use Polya's theorem with weights) as follows: Given a set $X=\{1,\dots,n\}$ of $n$ vertices (think about it as a necklace) and $C$ a set of $n$ colors with weights assigned to each color, i.e w(c_i)=x_i. Count the number of ways to do an n-coloring of X if two coloring are indistinguishable if we can obtain one just by swapping colors on the other. Since the group action will act on the set of colors (instead of $X$) I don't know how to follow... 
Any help will be appreciated! Thank you in advance.

I'm not sure why you would want to use unlabeled species here. It appears that the standard procedure is to observe that the OGF of the cycle index of the symmetric group is
$$G(z) = \sum_{q\ge 0} Z(S_q) z^q =
\exp\left(a_1 z + a_2 \frac{z^2}{2} + a_3 \frac{z^3}{3} + \cdots\right).$$
We are interested in the coefficient
$$n! [z^n a_1^{k_1} a_2^{k_2} a_3^{k_3}\cdots] G(z)$$
where $$\sum_{q=1}^n k_q q = n.$$
Actually doing the extraction we use
$$G(z) = \exp\left( \sum_{q\ge 1} a_q \frac{z^q}{q}\right)
= \prod_{q\ge 1}\exp\left(a_q \frac{z^q}{q}\right)$$
to obtain
$$n! [z^n] \prod_{q\ge 1} \frac{z^{q k_q}}{q^{k_q} k_q!}
= n! \prod_{q\ge 1} \frac{1}{q^{k_q} k_q!}.$$
