Example 7 Choose several colors from six different given colors to color the six faces of a cube, with each face being colored with exactly one color, and any two faces sharing a common edge must be colored with different colors. How many different coloring schemes are there?
(Note: If we color two identical cubes and can make the colors of the corresponding top, bottom, left, right, front, and back faces of the two cubes the same through appropriate rotation, then we say that the two cubes have the same coloring scheme).
Solution: Since two faces sharing a common vertex must be of different colors, at least 3 colors are required. Below, we divide the problem into 4 cases.
(1) When all 6 colors are used, first fix a face of a certain color facing up. There are $C$ ways to choose 1 color from the remaining 5 colors to paint the bottom face, and there are $(4-1)!$ ways to paint the remaining four side faces with the remaining 4 colors (a circular permutation of 4 colors). Therefore, when all 6 colors are used, there are $C \cdot (4-1)! = 30$ coloring schemes.
(2) When exactly 5 colors are used. There are $C_{6}^{5}$ ways to choose 5 colors from 6 colors. At this time, there must be a pair of opposite faces of the same color. Choose 1 color from the 5 colors to paint a pair of opposite faces, and place them facing up and down, with $C_{5}^{1}$ ways. The remaining 4 colors are used to paint the remaining four side faces (a necklace permutation of 4 different colored beads) with $\frac{(4-1)!}{2}$ ways. Therefore, when exactly 5 colors are used, there are
$$
C_{6}^{5} C_{5}^{1} \cdot \frac{1}{2}(4-1)! = 90 \text{ ways.}
$$
(3) When exactly 4 colors are used, there are $C_{6}^{4}$ ways to choose 4 colors from 6 colors. At this time, there must be 2 pairs of opposite faces of the same color, and another pair of opposite faces of different colors. Place the pair of different colors facing up and down, and choose 2 colors from the 4 colors to paint the top and bottom faces (at this time, the top and bottom faces are indistinguishable) with $C_{4}^{2}$ ways. The remaining 2 colors are used to paint the remaining four side faces so that the 2 pairs of opposite faces are of the same color (this can be seen as a necklace permutation of 2 colored beads, or a circular permutation) with only 1 way. Therefore, when exactly 4 colors are used, there are $C_{6}^{4} C_{4}^{2} \cdot 1 = 90$ coloring schemes.
(4) When exactly 3 colors are used, there are $C_{6}^{3}$ ways to choose 3 colors from 6 colors. At this time, all 3 pairs of opposite faces are of the same color, and there is only 1 way to paint them with 3 colors. Therefore, when exactly 3 colors are used, there are $C_{6}^{3} = 20$ coloring schemes.
In summary, there are a total of $30 + 90 + 90 + 20 = 230$ different coloring schemes.
/Comment: This problem emphasizes the counting of coloring schemes for a 3D figure when it is flipped. For such 3D figures, fixing some elements and converting the problem into a coloring counting problem for a 2D figure makes the solution clear. In case (2), the remaining 4 colors painting the four faces is equivalent to a necklace permutation of 4 different colored beads, not a circular permutation problem, which we should pay attention to. The reference answer given for this 1996 National High School Mathematics Competition problem was initially incorrect here, mistakenly calculating the final result as $30 + 180 + 90 + 20 = 320$ ways.