Example 4-10 In Figure 4-12(a), $v_{1} v_{2} v_{3}$ are 3 equally divided points on a circle, decorated with beads of red, blue, and green. How many different schemes are there?
From Fig. 4-12(a), it can be rotated around the center of the circle by $0^{\circ}, 120^{\circ}, 240^{\circ}$, and flipped over the vertical line through vertex 1 perpendicular to the line connecting the other two vertices, forming the group:
$$
G=\left\{\left(v_{1}\right)\left(v_{2}\right)\left(v_{3}\right),\left(v_{1} v_{2} v_{3}\right),\left(v_{3} v_{2} v_{1}\right),\left(v_{1}\right)\left(v_{2} v_{3}\right),\left(v_{2}\right)\left(v_{1} v_{3}\right),\left(v_{3}\right)\left(v_{1} v_{2}\right)\right\}
$$

Therefore, the number of different schemes is
$$
m=\frac{1}{6} \times\left[3^{3}+2 \cdot 3+3 \cdot 3^{2}\right]=\frac{1}{6} \times[27+6+27]=10
$$

See Fig. 4-12(b) for the 10 schemes.