Generating function of colouring a dodecahedron Find the generating functions in variables $r,b,g$ for the number of ways to paint a dodecahedron red,blue and green. 
For example the coefficient of $r^5b^4g^3$ should be the number of ways to paint the dodecahedron with 5 red faces, 4 blue faces and 3 green faces.
I know that we should use Polya's enumeration theorem but I have no idea how to continue. Any solutions are appreciated. 
Also it is allowed to be that r=0,or b=0 or g=0
 A: Let $G$ be the symmetry group of rotations for the dodecahedron.
Following this answer, we know
its cycle index is equal to
$$Z(G) = \frac{1}{60}
\left( a_1^{12} + 24 a_1^2 a_5^2 + 20 a_3^4 + 15  a_2^6\right).$$
It tells us the symmetry group $G$ consists of $60$ permutations:


*

*$\phantom{0}1$  permutation which is a product of twelve $1$-cycles. i.e the identity.

*$24$ permutations which is a product of two $1$-cycles and two $5$-cycles.

*$20$ permutations which is a product of four $3$-cycles.

*$15$ permutations which is a product of six $2$-cycles.


Let's say we are going to color the faces of a dodecahedron with $n$ colors $c_1, c_2, \ldots, c_n$.
Burnside's Lemma
tell us the number of ways is equal to
$$\frac{1}{|G|}\sum_{g\in G} |X_g|$$
where $X_g$ is the set of colorings (by colors $c_1, c_2, \ldots c_n$ ) invariant under the action of $g$.
Consider the case $g$ is a product of $m$ disjoint cycles of length $\ell_1, \ell_2, \ldots, \ell_m$ with $\sum_{j=1}^m \ell_j = 12$.
If a coloring is invariant under $g$, then every face belongs to same cycle
need to have same color. The $j^{th}$ cycle of length $\ell_j$ will contribute a
factor $c_1^{\ell_j} + c_2^{\ell_j} + \cdots + c_n^{\ell_j}$ to the GF. 
This leads to one simplified variant of
Pólya enumeration Theorem:

To compute the generating function for $n$ colors $c_1, c_2, \cdots, c_n$,
  replace every appearance of $a_k$ in the cycle index by $c_1^k + c_2^k + \cdots + c_n^k$.

Apply these to the case of $3$ colors $r, g, b$, the generating function
we seek is given by
$$\frac{1}{60}
\left( (r+g+b)^{12} + 24 (r+g+b)^2 (r^5+g^5+b^5)^2 \\+ 20(r^3+g^3+b^3)^4 + 15  (r^2+g^2+b^2)^6\right)$$
