Example 4 (1996 National High School League Question) From the given 6 different colors, select several colors to paint the 6 faces of a cube, with each face being painted one color, and two faces sharing an edge must be different colors. The number of different coloring schemes is $\qquad$ (coloring schemes that can coincide by rolling and rotating are considered the same coloring scheme).
Solution: Fill in 230. Reason: Since the three faces sharing a common vertex must be of different colors, at least 3 colors are required. Below, we divide into four cases.
(1) When all 6 colors are used, fix one color on the top face, and choose 1 color from the remaining 5 colors to color the bottom face, which has $C$ ways. The remaining 4 colors are used to color the remaining 4 side faces (which should be a circular permutation of 4 colors) with $(4-1)!$ ways. Therefore, when all 6 colors are used, there are $C_{5}^{1} \cdot (4-1)! = 30$ coloring schemes.
(2) When only 5 colors are used, choose 5 colors from 6 colors, which has $C_{6}^{5}$ ways. At this time, there must be a pair of opposite faces with the same color. Choose 1 color from the 5 colors to color a pair of opposite faces, and place them on top and bottom, which has $C_{5}^{1}$ ways. The remaining 4 colors are used to color the remaining 4 side faces (which should be a necklace of 4 different colored beads) with $\frac{1}{2} \cdot (4-1)!$ ways. Therefore, when only 5 colors are used, there are $C_{6}^{5} \cdot C_{5}^{1} \cdot \frac{1}{2} \cdot (4-1)! = 90$ different coloring schemes.
(3) When only 4 colors are used, choose 4 colors from 6 colors, which has $C_{6}^{4}$ ways. At this time, there must be 2 pairs of opposite faces with the same color, and another pair of opposite faces with different colors. Place the pair of opposite faces with different colors on top and bottom, and choose 2 colors from the 4 colors to color the top and bottom faces (note, at this time the top and bottom faces are indistinguishable), which has $C_{4}^{2}$ ways. The remaining 2 colors are used to color the 4 side faces so that the 2 pairs of opposite faces have the same color (which should be a necklace of 2 colored beads) with only 1 way. Therefore, when only 4 colors are used, there are $C_{6}^{4} \cdot C_{4}^{2} \cdot 1 = 90$ different coloring schemes.
(4) When only 3 colors are used, choose 3 colors from 6 colors, which has $C_{6}^{3}$ ways. At this time, all 3 pairs of opposite faces have the same color, and using 3 colors to color them has only 1 way. Therefore, when only 3 colors are used, there are $C_{6}^{3} = 20$ different coloring schemes.
In summary, there are a total of $30 + 90 + 90 + 20 = 230$ different coloring schemes.