Example 4.4.4 A die has six faces marked with $1,2, \cdots, 6$, how many different dice are there?
Given: $D=\{1,2,3,4,5,6\}, R=\left\{w_{1}, w_{2}, \cdots, w_{6}\right\}$.
$D$'s permutation group $G$ has 24 elements (as in Example 4.4.3).
$$
P_{G}\left(x_{1}, x_{2}, x_{3}, x_{4}, x_{5}, x_{6}\right)=\frac{1}{24}\left(x_{1}^{6}+6 x_{1}^{2} x_{4}+3 x_{1}^{2} x_{2}^{2}+8 x_{3}^{2}+6 x_{2}^{3}\right).
$$

Thus,
$$
\begin{aligned}
F_{G} & \left(\sum_{i=1}^{6} w_{i}, \sum_{i=1}^{6} w_{i}^{2}, \cdots, \sum_{i=1}^{6} w_{i}^{6}\right) \\
= & \frac{1}{24}\left[\left(w_{1}+w_{2}+\cdots+w_{6}\right)^{6}+6\left(w_{1}+w_{2}+\cdots+w_{6}\right)^{2}\left(w_{1}^{4}+w_{2}^{4}\right.\right. \\
& \left.+\cdots+w_{6}^{4}\right)+3\left(w_{1}+w_{2}+\cdots+w_{6}\right)^{2}\left(w_{1}^{2}+w_{2}^{2}+\cdots+w_{6}^{2}\right)^{2}+8 \\
& \left.\left(w_{1}^{3}+w_{2}^{3}+\cdots+w_{6}^{3}\right)^{2}+6\left(w_{1}^{2}+w_{2}^{2}+\cdots+w_{6}^{2}\right)^{3}\right],
\end{aligned}
$$

The coefficient of $w_{1} w_{2} w_{3} w_{4} w_{5} w_{6}$ is the number of ways to color the 6 faces with 6 different colors, which is $\frac{1}{24} \cdot \frac{6!}{1!1!\cdots 1!}=30$.

Note: The result can also be calculated simply without using Pólya's enumeration theorem as follows: There are $6!$ ways to color the 6 faces with 6 different colors. However, among these, those that can be transformed into each other by rigid motions are considered the same. So, the 6! includes repeated counts. The bottom face can be colored in 6 ways, and after fixing the color of the bottom face, there are 4 rotations that make the schemes equivalent. Therefore, each coloring scheme has 24 equivalent schemes, and the repetition factor in 6! is 24. Thus, the number of distinct coloring schemes is $\frac{6!}{24}=30$.