1.107 From the given 6 different colors, select several colors to paint the 6 faces of a cube, with each face painted one color, and any two adjacent faces sharing an edge must be painted different colors. How many different painting schemes are there? (If one of the two painted cubes can be made to have exactly the same coloring as the other by several rotations, then the original coloring schemes of the two are considered the same)


[Solution 1] (1) When using 6 colors, the first color can be placed on the top, and one of the 4 side colors can be placed forward. Thus, the number of different coloring schemes is
$$
6! \div 6 \div 4 = 30.
$$
(2) When using 5 colors, choose 5 colors from 6, and select 1 color to paint the two opposite faces, say the top and bottom. Then choose one of the remaining 4 colors to face forward, and the back will be one of the remaining 3 colors. The remaining two colors can be painted on the left and right faces, which can be interchanged by flipping. Therefore, the number of different coloring schemes is
$$
C_{6}^{5} C_{5}^{1} \times 3 = 6 \times 5 \times 3 = 90.
$$
(3) When using 4 colors, choose 4 colors from 6, and select 2 colors to paint the two opposite faces. The number of different coloring schemes is
$$
C_{6}^{4} C_{4}^{2} = 15 \times 6 = 90.
$$
(4) The number of different coloring schemes using 3 colors is $C_{6}^{3} = 20$.

In summary, the total number of different coloring schemes is $30 + 90 + 90 + 20 = 230$.
[Solution 2] (1) When using 6 colors, the first color can always be placed on the bottom, and one of the remaining 5 colors can be chosen to paint the top, giving 5 choices; from the remaining 4 colors, one can be designated to paint the back, and one of the remaining 3 colors can be chosen to paint the front, giving 3 choices; the remaining two colors can be painted on the left and right faces, giving 2 choices. There are a total of 30 choices.
(2) When using 5 colors, exactly one color is used to paint two faces. Clearly, these must be two opposite faces, which can be the top and bottom. This gives 6 choices; from the remaining 5 colors, 4 are chosen to paint the other faces. Choosing 4 colors gives 5 choices, and from these 4 colors, one is designated to paint the back, and one of the remaining 3 colors is chosen to paint the front, giving 3 choices; the remaining two colors can be painted on the left and right faces, which can be interchanged. By the multiplication principle, there are $6 \times 5 \times 3 = 90$ coloring schemes.
(3) When using 4 colors, exactly two colors are used to paint two pairs of opposite faces. By symmetry, the other two colors have only 1 way to be painted, so the number of different coloring schemes is $C_{6}^{4} C_{4}^{2} = 90$.
(4) When using 3 colors, by symmetry, there are $C_{6}^{3} = 20$ ways.
In summary, the total number of different coloring schemes that meet the requirements of the problem is $30 + 90 + 90 + 20 = 230$.