Example 4-16 There are 3 different colored beads, used to make a necklace of 4 beads, what are the possible schemes?

 

Using 3 different colored beads to assemble a necklace of 4 beads, the question asks for all possible combinations.
As shown in Figure 4-17, the movements that make them coincide include rotations of $\pm 90^{\circ}$ and $180^{\circ}$ about the center of the ring, and flips of $180^{\circ}$ about the $x x^{\prime}$ and $y y^{\prime}$ axes. Therefore, the permutation group $G$ is
$$
\begin{array}{c}
\left(v_{1}\right)\left(v_{2}\right)\left(v_{3}\right)\left(v_{4}\right),\left(v_{2}\right)\left(v_{4}\right)\left(v_{1} v_{3}\right), \\
\left(v_{1} v_{2} v_{3} v_{4}\right),\left(v_{1}\right)\left(v_{3}\right)\left(v_{2}\right)\left(v_{4}\right), \\
\left(v_{1} v_{3}\right)\left(v_{2} v_{4}\right),\left(v_{1} v_{4}\right)\left(v_{2} v_{3}\right), \\
\left(v_{1} v_{3} v_{2} v_{1}\right),\left(v_{1} v_{2}\right)\left(v_{2} v_{4}\right)
\end{array}
$$
The number of cases is
$$
m=\frac{1}{8} \times\left(3^{4}+2 \cdot 3+3 \cdot 3^{2}+2 \cdot 3^{3}\right)=21
$$

The specific schemes are
$$
\begin{aligned}
P= & \frac{1}{8} \times\left[(b+g+r)^{4}+2\left(b^{4}+g^{4}+r^{4}\right)+3\left(b^{2}+g^{2}+r^{2}\right)^{2}\right. \\
& \left.+2\left(b^{2}+g^{2}+r^{2}\right)(b+g+r)^{2}\right] \\
= & b^{4}+r^{4}+g^{4}+b^{3} r+b^{3} g+b r^{3}+r g+b g^{3} \\
& +r g^{3}+2 b^{2} r^{2}+2 b^{2} g^{2}+2 r^{2} g^{2}+2 b^{2} r g+2 b r^{2} g+2 b r g^{2}
\end{aligned}
$$

The coefficient of $b^{2} r g$ is 2, meaning there are two schemes with two blue beads, one red bead, and one green bead, as shown in Figure 4-18.