Example 4-15 The 6 faces of a die have $1,2,3,4,5,6$ dots respectively, how many different schemes are there?
Solution I:
The problem is equivalent to coloring the 6 faces of a cube with 6 different colors, requiring that each face has a different color, and finding the number of different schemes.
There should be $6!$ schemes for coloring 6 faces with 6 different colors, but among them, those that can be superimposed by rigid motion are considered the same. From Example 4-13, we know that the motion transformation group has 24 elements. Below, we observe the effect of the group elements on the $6!$ schemes.
Let the $6!$ schemes be $S_{1}, S_{2}, \cdots, S_{6!}$.
Under the action of the identity element, there is the permutation
$$
\left(S_{1}\right)\left(S_{2}\right) \cdots\left(S_{6!}\right)
$$

Since the 6 faces have different colors, none of the other 23 permutations can leave any scheme unchanged. Let $p_{0}$ be the identity element, i.e., the identity permutation, and the rest be $p_{1}, p_{2}, \cdots, p_{23}$. Therefore, we have
$$
\begin{array}{l}
c_{1}\left(p_{0}\right)=6!, \\
c_{1}\left(p_{1}\right)=c_{1}\left(p_{2}\right)=c_{1}\left(p_{3}\right)=\cdots=c_{1}\left(p_{23}\right)=0
\end{array}
$$

According to Burnside's Lemma, the number of different schemes should be
$$
m=\frac{1}{24} \times[6!+0+\cdots+0]=30
$$

Solution 2:
The following uses Pólya's Theorem and the principle of inclusion-exclusion.
The permutation group that makes the cube coincide is seen in Example 4-13. Using $m$ colors to paint the 6 faces of the cube, the number of different schemes is $n_{m}$, given by
$$
n_{m}=\frac{1}{24} \times\left[m^{6}+3 m^{4}+12 m^{3}+8 m^{2}\right]
$$

Then we have
$$
\begin{array}{l}
n_{1}=\frac{1}{24} \times[1+3+12+8]=\frac{24}{24}=1 \\
n_{2}=\frac{1}{24} \times\left[2^{6}+3 \cdot 2^{4}+12 \cdot 2^{3}+8 \cdot 2^{2}\right]=\frac{1}{24} \times[240]=10 \\
n_{3}=\frac{1}{24} \times\left[3^{6}+3 \cdot 3^{4}+12 \cdot 3^{3}+8 \cdot 3^{2}\right]=\frac{1}{24} \times[1368]=57 \\
n_{4}=\frac{1}{24} \times\left[4^{6}+3 \cdot 4^{4}+12 \cdot 4^{3}+8 \cdot 4^{2}\right]=\frac{1}{24} \times[5760]=240 \\
n_{5}=\frac{1}{24} \times\left[5^{6}+3 \cdot 5^{4}+12 \cdot 5^{3}+8 \cdot 5^{2}\right]=\frac{1}{24} \times[19200]=800 \\
n_{6}=\frac{1}{24} \times\left[6^{6}+3 \cdot 6^{4}+12 \cdot 6^{3}+8 \cdot 6^{2}\right]=\frac{1}{24} \times[53424]=2226
\end{array}
$$

Let $l_{i}=$ the number of schemes using $i$ colors to paint the 6 faces of the cube, such that there are at least $i$ different colors, then we have
$$
\begin{aligned}
l_{1} & =n_{1}=1 \\
l_{2} & =n_{2}-\binom{2}{1} l_{1}=n_{2}-2 l_{1}=10-2=8 \\
l_{3} & =n_{3}-\binom{3}{2} l_{2}-\binom{3}{1} l_{1}=n_{3}-3 l_{2}-3 l_{1}=57-3 \times 8-3=30 \\
l_{4} & =n_{4}-\binom{4}{3} l_{3}-\binom{4}{2} l_{2}-\binom{4}{1} l_{1}=240-120-48-4=68 \\
l_{5} & =n_{5}-\binom{5}{4} l_{4}-\binom{5}{3} l_{3}-\binom{5}{2} l_{2}-5 l_{1}=800-340-300-80-5=75 \\
l_{6} & =n_{6}-\binom{6}{5} n_{5}-\binom{6}{4} n_{4}-\binom{6}{3} l_{3}-\binom{6}{2} l_{2}-6 l_{1} \\
& =2226-6 \times 75-15 \times 68-20 \times 30-15 \times 8-6=30
\end{aligned}
$$