Example 4-18 Embed 4 red beads in 4 corners of a cube, try to find how many schemes there are?
The problem is equivalent to coloring the vertices of a regular hexahedron with two colors, and finding the number of schemes where the two colors are equal. From Example 4-14 in Section 4.6, we know that the order of the permutation group is $=24$, with 1 of format $(1)^{8}$, 6 of format $(4)^{2}$, 9 of format $(2)^{4}$, and 8 of format $(1)^{2}(3)^{2}$.
$$
P=\frac{1}{24} \times\left[(b+r)^{8}+6\left(b^{4}+r^{4}\right)^{2}+9\left(b^{2}+r^{2}\right)^{4}+8(b+r)^{2}\left(b^{3}+r^{3}\right)^{2}\right]
$$

The coefficient of $b^{4} r^{4}$ is
$$
\begin{aligned}
& \frac{1}{24} \times[C(8,4)+12+9 C(4,2)+8 C(2,1) C(2,1)] \\
= & \frac{1}{24}\left[\frac{8 \times 7 \times 6 \times 5}{4!}+12+9 \frac{4 \times 3}{2!}+32\right]=7
\end{aligned}
$$

The corresponding schemes are shown in Figure 4-20.