Example 4.4.1 Using red, blue, and green to color the vertices of a regular pentagon, how many inequivalent methods are there?
Solution: $D=\{1,2,3,4,5\}, R=\{r, b, g\}, C$ has a total of $3^{5}=243$ coloring methods. Let $P_{5}$ be a rotation of $\frac{360^{\circ}}{5}$ around the center, $e$ be the identity permutation, and $t_{i}$ be the reflection over the line connecting vertex $i$ and the midpoint of its opposite edge. Then the permutation group $G=\left\{e, P_{5}, P_{5}^{2}, P_{5}^{3}, P_{5}^{4}, t_{1}, t_{2}, t_{3}, t_{4}, t_{5}\right\}$.

To calculate $N(G, C)$, we must find $|C(\sigma)|$ for each permutation. For the identity permutation $e$, every coloring scheme remains unchanged, i.e., $C(e)=243$. Each rotation only keeps three coloring schemes (the three schemes where all five points are the same color) unchanged, i.e., $C\left(P_{5}^{i}\right)=3$. Each reflection keeps $3 \times 3 \times 3=27$ coloring schemes unchanged (for example, under $t_{1}$, the unchanged colorings must have 2 and 5 the same color, 3 and 4 the same color, and 1 can be any color, hence there are $3 \times 3 \times 3$ choices), i.e., $C\left(t_{i}\right)=27$. Therefore, $N(G, C)=\frac{1}{10}(243+3 \times 4+27 \times 5)=39$.
5. Pólya Counting Formula for the Number of Non-Equivalent Colorings
Let $G$ be a permutation group of $D$. For $\sigma \in G$, $\sigma$ is of the type $1^{b_{1}} 2^{b_{2}} \cdots n^{b_{n}}$.
For $\sigma$, form the monomial $\operatorname{mon}(\sigma)=x_{1}^{b_{1}} x_{2}^{b_{2}} \cdots x_{n}^{b_{n}}$, and sum the monomials of each permutation $\sigma$ in $G$, to obtain the generating function of $G$ according to the type
$$
P_{G}\left(x_{1}, x_{2}, \cdots, x_{n}\right)=\frac{1}{|G|} \sum_{\sigma \in G} \operatorname{mon}(\sigma)=\frac{1}{|G|} \sum_{\sigma \in G} x_{1}^{b_{1}} x_{2}^{b_{2}^{2}} \cdots x_{n}^{b_{n}} .
$$

Pólya Counting Theorem: Let $D=\{1,2, \cdots, n\}$ be a set of $n$ distinct elements, and let $C$ be the set of all $m^{n}$ colorings of $D$ using $m$ colors. Let $G$ be a permutation group of $D$. Then the number of non-equivalent colorings is
$$
N(G, C)=P_{G}(m, m, \cdots, m) .
$$

where $P_{G}\left(x_{1}, x_{2}, \cdots, x_{n}\right)$ is the generating function of the permutation group $G$.
Let $\left\{w_{1}, w_{2}, \cdots, w_{m}\right\}$ be the set of $m$ colors, then
$$
P_{G}\left(\sum_{i=1}^{m} w_{i}, \sum_{i=1}^{m} w_{i}^{2}, \cdots, \sum_{i=1}^{m} w_{i}^{n}\right)
$$

is the generating function for the number of non-equivalent colorings. The coefficient of $w_{1}^{P_{1}} w_{2}^{P_{2}} \cdots w_{n}^{P_{n}}$ in the expansion is the number of non-equivalent colorings in $C$ where $P_{1}$ elements are colored $w_{1}$, $P_{2}$ elements are colored $w_{2}$, $\cdots P_{n}$ elements are colored $w_{n}$.