10,11
a) Six different colors are chosen; it is required to paint 6 faces of a cube, each in a special color from the chosen ones. In how many geometrically distinct ways can this be done? Geometrically distinct are two such colorings that cannot be made to coincide with each other by rotating the cube around its center.
b) Solve the same problem for the case of coloring the faces of a dodecahedron in 12 different colors.
a) The first method. Suppose the procedure of painting a cube goes as follows: an unpainted cube is placed in a machine in a certain fixed position, and then its faces are painted in a certain order. There are as many such ways as there are permutations of 6 colors, that is, 6!. However, the cube can be placed in a fixed position in 24 different ways: the cube can be placed on any of the 6 faces and then rotated around the vertical axis in any of four ways. Therefore, the number of geometrically distinct colorings is 24 times less, that is, $6! / 24 = 30$.

The second method. A cube can always be rotated so that a face of the desired (say, white) color is at the bottom, so we can assume that the bottom face is always painted white. After this, there are 5 ways to choose a color for the opposite face. From the remaining four colors, we fix one and paint the front face with it (other coloring options can be ignored since the cube can always be rotated around the vertical axis to such a position). There are 3! ways to paint the three remaining faces. In total, we get $5 \cdot 3! = 30$ ways.

b) The reasoning is completely analogous to a) (first method).

The total number of colorings is 12!. A dodecahedron can be placed in a fixed position in 60 ways: by placing it on any of the 12 faces and then rotating it in one of five ways. Therefore, the answer is $12! / 60$.

## Answer

a) 30; b) $12! / 60 = 11! / 5$ colorings.