11. From the given six different colors, select several colors to paint the six faces of a cube, with each face painted exactly one color, and any two faces sharing an edge must be painted different colors. Then, the number of different coloring schemes is $\qquad$. (Note: If we can, by appropriate rotation, make the top, bottom, left, right, front, and back faces of two identically painted cubes match in color, we say that the two cubes have the same coloring scheme.)
Using 6 colors: the top is a fixed color, the bottom has 5 choices, and the remaining 4 faces have $(4-1)!=6$ choices, totaling 30 coloring schemes; Using 5 colors: the top and bottom use the same color, with 6 choices, the remaining 4 faces have $C_{5}^{4}(4-1)!=30$ choices, considering flipping, there are $\frac{6 \times 30}{2}=90$ coloring schemes; Using 4 colors: there are $C_{6}^{4} \cdot C_{4}^{2} \cdot \frac{2!}{2}=90$ coloring schemes; Using 3 colors: there are $C_{6}^{3}=20$ coloring schemes. Therefore, the total number of coloring schemes is 230.