3-ча 1. 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.

Solve the same problem for the case of coloring the faces of a regular dodecahedron in twelve different colors.

(This problem was not solved by any of the participants in the olympiad.)
Solution 1. A cube can be rotated so that the face painted the first color takes the specified position. There are 5 different options for painting the opposite face; different colorings of the opposite face give geometrically different colorings of the cube.

Among the remaining four faces, one can choose the face painted a given color and translate it to a given position (without changing the position of the first two faces). Different colorings of the three remaining faces give geometrically different colorings of the cube. One of these faces can be painted in 3 ways, and one of the remaining in 2 ways. In total, we get $5 \cdot 3 \cdot 2=30$ geometrically different colorings.

Now let's solve the problem for a 12-faced polyhedron (dodecahedron). The total number of possible colorings of the dodecahedron is $12!=1 \cdot 2 \cdots \cdots 12$. To find the number of geometrically different colorings, we need to divide 12! by the number of self-coincidences of the dodecahedron. Any of the 12 faces can be translated to any other. In addition, there are 5 rotations (including the identity) that preserve a given face. In total, there are 60 self-coincidences. Therefore, the number of geometrically different colorings of the dodecahedron is $12!/ 60=7983360$.