Example 7.18 Use red, pink, yellow, and black to paint the 6 faces of a cube, so that the faces painted yellow and black each have one, and the faces painted red and pink each have two. Mr. Li is somewhat color blind. For two colored cubes $W_{6}$ and $W_{6}^{\prime}$, if $W_{6}^{\prime}$ is obtained by changing the 2 red faces of $W_{6}$ to pink, and the 2 pink faces to red, Mr. Li, due to his color blindness, simply considers $W_{6}$ and $W_{6}^{\prime}$ to be the same. Find the number of distinct coloring patterns of the cube under Mr. Li's identification method.
Let $A$ denote the set of the 6 faces of a cube, and let $b_{1}, b_{2}, b_{3}, b_{3}^{\prime}, b_{4}, b_{4}^{\prime}$ represent 6 different colors. Let $B=\left\{b_{1}, b_{2}, b_{3}, b_{3}^{\prime}, b_{4}, b_{4}^{\prime}\right\}$. Let $(D, *)$ denote the rotation group of the cube, and $(G, \circ)$ be the face permutation group induced by $(D, *)$; let $(H, \cdot)$ denote the permutation group on the set $B$ as follows: $H=\left\{h_{1}, h_{2}, \cdots, h_{8}\right\}$, where
$$
\begin{array}{l}
h_{1}=\left(b_{1}\right)\left(b_{2}\right)\left(b_{3}\right)\left(b_{3}^{\prime}\right)\left(b_{4}\right)\left(b_{4}^{\prime}\right) \\
h_{2}=\left(b_{1}\right)\left(b_{2}\right)\left(b_{3} b_{3}^{\prime}\right)\left(b_{4}\right)\left(b_{4}^{\prime}\right) ; \\
h_{3}=\left(b_{1}\right)\left(b_{2}\right)\left(b_{3}\right)\left(b_{3}^{\prime}\right)\left(b_{4} b_{4}^{\prime}\right) \\
h_{4}=\left(b_{1}\right)\left(b_{2}\right)\left(b_{3} b_{3}^{\prime}\right)\left(b_{4} b_{4}^{\prime}\right) ; \\
h_{5}=\left(b_{1}\right)\left(b_{2}\right)\left(b_{3} b_{4}\right)\left(b_{3}^{\prime} b_{4}^{\prime}\right) ; \\
h_{6}=\left(b_{1}\right)\left(b_{2}\right)\left(b_{3} b_{4}^{\prime}\right)\left(b_{3}^{\prime} b_{4}\right) ; \\
h_{7}=\left(b_{1}\right)\left(b_{2}\right)\left(b_{3} b_{4} b_{3}^{\prime} b_{4}^{\prime}\right) \\
h_{8}=\left(b_{1}\right)\left(b_{2}\right)\left(b_{3} b_{4}^{\prime} b_{3}^{\prime} b_{4}\right)
\end{array}
$$

Then the cycle index of $(G, \circ)$ is
$$
P_{G}\left(x_{1}, x_{2}, \cdots, x_{6}\right)=\frac{1}{24}\left(x_{1}^{6}+8 x_{3}^{2}+6 x_{2}^{3}+3 x_{1}^{2} x_{2}^{2}+6 x_{1}^{2} x_{4}\right) \text {, }
$$
and the cycle index of $(H, \cdot)$ is
$$
P_{H}\left(x_{1}, x_{2}, \cdots, x_{6}\right)=\frac{1}{8}\left(x_{1}^{6}+2 x_{1}^{4} x_{2}+3 x_{1}^{2} x_{2}^{2}+2 x_{1}^{2} x_{4}\right) .
$$

By a discussion similar to Example 7.17, it is easy to see that the number of distinct colored cubes $N$ is equal to the number of $(G, H)$-orbits of the set of all bijections from $B$ to $A$. By Theorem 7.35, we have
$$
\begin{aligned}
N= & \frac{1}{24 \times 8}\left[\left(\left(\frac{\partial}{\partial y_{1}}\right)^{6}+8\left(\frac{\partial}{\partial y_{3}}\right)^{2}+6\left(\frac{\partial}{\partial y_{2}}\right)^{3}+\right.\right. \\
& \left.3\left(\frac{\partial}{\partial y_{1}}\right)^{2}\left(\frac{\partial}{\partial y_{2}}\right)^{2}+6\left(\frac{\partial}{\partial y_{1}}\right)^{2}\left(\frac{\partial}{\partial y_{4}}\right)\right)\left(y_{1}^{6}+2 y_{1}^{4}\left(2 y_{2}\right)+\right. \\
& \left.\left.3 y_{1}^{2}\left(2 y_{2}\right)^{2}+2 y_{1}^{2}\left(4 y_{4}\right)\right)\right]_{y_{1}=y_{2}=\cdots=y_{6}=0} \\
= & \frac{1}{192}\left(6!+3 \times 3 \times 2!\times 2^{2} \times 2!+6 \times 2 \times 2!\times 4\right) \\
= & 5 .
\end{aligned}
$$