Let $S_{n}$ be the symmetric group of an $n$-element set $D$ and $C$ be a set of $m$ elements with weights $x_{1},x_{2},\dots,x_{m}$. Two functions $f_{1},f_{2}\in C^D$ are $S_{n}$-equivalent if and only if they have the same weight: if they use the each colour the same number of times, then clearly some permutation of $D$ sends $f_{1}$ to $f_{2}$. That is, for any possible function weight $\omega$, $S_{\omega}$ as defined earlier contains one orbit under the action of $S_{n}$. Therefore $\sum_{F\in C^D/S_{n}} W(F) = \sum_{i_{1}+i_{2}+\dots + i_{m}=n} x_{1}^{i_{1}} x_{2}^{i_{2}} \dots x_{m}^{i_{m}}.$While we can compute this directly without referencing the cycle index $Z_{S_n}$, it is still interesting to investigate whether there is a some nice formula for this polynomial. Notice that $\sum_{n \geq 0} \left( \sum_{F\in C^D/S_{n}} W(F) \right) y^n = \frac{1}{(1-x_{1}y)(1-x_{2}y)\cdots(1-x_{m}y)}$since the right-hand side is $\prod_{i=1}^{m}(1+x_{i}y+x_{i}^2y^2+\cdots)$ and the coefficient of $y^n$ in its expansion is the sum of all monomials $x_1^{i_1} x_2^{i_2} \dots x_m^{i_m}$ such that $i_1 + i_2 + \dots + i_m = n$, which matches the left-hand side. Since $-\ln(1-x)=x+\frac{x^2}{2}+\frac{x^3}{3}+\cdots$we have $\begin{align} \prod_{i=1}^{m} \frac{1}{(1-x_{i}y)} &= \prod_{i=1}^{m} \exp (-\ln(1-x_{i}y)) \\ &= \prod_{i=1}^{m}\exp \left( x_{i}y + \frac{x_{i}^{2}y^{2}}{2} + \frac{x_{i}^{3}y^{3}}{3} + \dots \right) \\ &= \exp \left( y \sum_{i=1}^{m} x_{i} + \frac{y^{2}}{2} \sum_{i=1}^{m} x_{i}^2 + \frac{y^{3}}{3} \sum_{i=1}^{m} x_{i}^{3}+ \dots \right) \end{align}$which we rewrite as $\exp\left( yz_{1}+ \frac{y^{2}}{2} z_{2} + \frac{y^{3}}{3}z_{3} +\dots \right)$where $z_{j}=\sum_{i=1}^{m}x_{i}^j$ for $j=1,2,\dots,n$. Recall also that Pólya's theorem gives $\sum_{F\in C^D/S_{n}} W(F)=Z_{S_{n}}(z_{1},z_{2},\dots, z_{n}).$Thus $\sum_{n \geq 0 } Z_{S_{n}}(z_{1},z_{2},\dots,z_{n})y^{n}= \exp\left( yz_{1}+ \frac{y^{2}}{2} z_{2} + \frac{y^{3}}{3}z_{3} +\dots \right)$or we may write equivalently $Z_{S_{n}}(z_{1},z_{2},\dots,z_{n})= [y^n]\exp\left( yz_{1}+ \frac{y^{2}}{2} z_{2} + \frac{y^{3}}{3}z_{3} +\dots \right).$ Furthermore, by observing that $\begin{align} \exp\left( yz_{1}+ \frac{y^{2}}{2} z_{2} + \frac{y^{3}}{3}z_{3} +\dots \right) &= \exp(yz_{1})\cdot\exp\left( \frac{y^{2}}{2} z_{2} \right)\cdot\exp\left( \frac{y^{3}}{3}z_{3} \right)\cdots \\ &= \left( \sum_{i\geq 0} \frac{y^{i}z_{1}^{i}}{i!} \right) \left( \sum_{i\geq 0} \frac{y^{2i}z_{2}^{i}}{2^{i}i!} \right) \left( \sum_{i\geq 0} \frac{y^{3i}z_{3}^{i}}{3^{i}i!} \right) \cdots, \end{align}$we derive that the coefficient of $z_{1}^{i_{1}}z_{2}^{i_{2}}\cdots z_{n}^{i_{n}}$ in $n! Z_{S_{n}}(z_{1},\dots,z_{n})$ (i.e. the number of permutations in $S_{n}$ with cycle type $(i_{1},i_{2},\dots ,i_{n})$) is equal to $\frac{n!}{i_{1}!i_{2}!\cdots i_{n}!\cdot 1^{i_{1}} 2^{i_{2}} \cdots n^{i_{n}} }.$This can also be argued combinatorially (see ^[1, Theorem 6.9]) We consider another application of this formula. Let $f(n)$ be the number of permutations in the symmetric group $S_{n}$ whose cycles have even length so that $f(n)=n! Z_{S_{n}}(z_{i}=0, \;i \,\text{ even}; z_{i}=1, \; i \; \text{odd}).$Then substituting gives $\sum_{n\geq 0} f(n) \frac{x^{n}}{n!} = \exp\left( \frac{x^{2}}{2} + \frac{x^{4}}{4} + \dots \right).$ Since $-\ln(1-x)=x+\frac{x^2}{2}+\frac{x^3}{3}+\cdots$, we get $\begin{align} \sum_{n\geq 0} f(n) \frac{x^{n}}{n!} &= \exp\left( \frac{1}{2} ( -\ln(1-x) - \ln(1+x) ) \right) \\ &= \exp \left( -\frac{1}{2} \ln(1-x^{2}) \right)\\ &= \left( 1-x^2 \right)^{-1/2}. \end{align}$Now the power series expansion of $\left( 1-x^2 \right)^{-1/2}$ centred at $0$ is given by $\left( 1-x^2 \right)^{-1/2}= \sum_{m \geq 0} (-1)^m{-1/2 \choose m}x^{2m}$where ${-1/2 \choose m}=\frac{1}{m!}\left( -\frac{1}{2} \right)\left( -\frac{3}{2} \right)\dots\left( -\frac{2m-1}{2} \right).$ Thus $f(n) = \begin{cases} 1^{2}\cdot 3^{2} \cdot 5^{2} \cdots (n-1)^{2}, & \text{if $n$ is even,} \\ 0, &\text{if $n$ is odd}, \end{cases}\tag{1}$(see \[OEIS A001818\]). We can show similarly that $g(n) = \begin{cases} 1^{2}\cdot 3^{2} \cdot 5^{2} \cdots (n-1)^{2}, & \text{if $n$ is even,} \\ 1^{2}\cdot 3^{2} \cdot 5^{2} \cdots(n-2)^{2}\cdot n, &\text{if $n$ is odd}. \end{cases}\tag{2}$where $g(n)$ is the number of the permutations in $S_{n}$ with odd order noting that a permutation has odd order if and only if all its cycles have odd length since the order of the permutation is the lowest common multiple of the lengths of its cycles (see \[OEIS A000246\]). Bóna ^[1, Theorem 6.24] provides a combinatorial proof of $(1)$ by considering how to construct such permutations. He continues to prove $(2)$ ^[1, Theorem 6.25] by establishing a bijection between the set of permutations in $S_{2n}$ where all cycles have odd length, together with a choice of one of $2n+1$ possible positions, and the set of permutations in $S_{2n+1}$ where all cycles have odd length. # References ^[1] M. Bóna, A walk through combinatorics, World Scientific Publishing Co. Pte. Ltd., Hackensack, NJ, 2006.