Final answer:
There are 5040 distinguishable ways to paint the faces of a regular hexagonal prism with 8 different colors so that no color is used more than once. This solution is derived by considering the symmetry of the hexagonal prism and applying the concept of permutations.
Step-by-step explanation:
The question asks how many distinguishable ways the faces of a regular hexagonal prism can be painted with 8 different colors, using one color per face, without using any color more than once. In geometry, particularly in the context of combinatorics, this type of problem is known as a permutation problem since we are interested in arranging the different colors on the different faces.
First, we must understand the structure of a hexagonal prism. It has two hexagonal faces and six rectangular faces between them. As we are required to use 8 colors and there are only 8 faces, each face will be painted a unique color. We tackle this question by considering the rotational symmetry of the hexagonal prism. The hexagonal faces cannot be distinguished from each other by rotation, and the same goes for the six rectangular faces.
As a result, we have two hexagonal faces where the order in which we paint them does not matter, and then six other faces to consider. To calculate the number of distinguishable ways, we can use the formula for permutations of different things taken all at a time, which is n!, where n is the number of things to arrange. So we have to arrange 8 colors on 8 faces, which gives us 8! (8 factorial) ways, but we need to divide by the number of indistinguishable arrangements due to the prism's symmetry.
There are 6 faces which form a cycle, and any rotation of these faces gives us an indistinguishable arrangement. There are 6 such rotations, hence the number of distinguishable permutations is 8! divided by 6. This gives us:
8! / 6 = (8 × 7 × 6 × 5 × 4 × 3 × 2 × 1) / 6 = 7 × 5 × 4 × 3 × 2 × 1 = 7! = 5040 distinguishable ways.