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Use Polya's enumeration to determine the number of six-sided dice that can be manufactured if each of three different labels must be placed on two of the faces.

Can you help me please to solve this exercise ? I have looked for Polya's enumeration formula on Wikipedia but it is very strange for me.

Thanks :)

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This blog post might be helpful. It has an example involving labeling squares, and then one about labeling cubes. –  MJD Jan 4 '13 at 20:26
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I think that "each of three different labels must be placed on two of the faces" means that there are two red, two blue, and two yellow stickers on the six faces. I hope that's right.

First you enumerate the elements of the symmetry group of the cube. There are 24 of these, falling into five different classes:

  1. 6 elements that are rotations of $\pm90^\circ$ around axes through face centers
  2. 3 elements that are rotations of $180^\circ$ around axes through face centers
  3. 8 elements that are rotations of $\pm120^\circ$ around axes through opposite vertices
  4. 6 elements that are rotations of $180^\circ$ around axes through the midpoints of opposite edges
  5. 1 identity element

Our job is to count how many ways there are to assign colors to the faces that are left fixed by each of the 24 elements. The average of these 24 counts is the answer we want.

Each of these 5 sorts of symmetries divides the faces of the cube into "orbits", which are equivalence classes of faces which can be mapped to one another by just that symmetry. All the faces in a single orbit must receive the same color sticker. For elements of sort 1, there are three orbits, consisting of 1, 1, and 4 faces respectively; since we don't have 4 stickers of the same color, there are no colorings left fixed by this sort of symmetry.

For elements of sort 2, there are four orbits of sizes 1, 1, 2, and 2 faces. The 2-face orbits each get one color and the two 1-face orbits get the last color. There are 6 ways to assign the 3 colors to the orbits, so each of these 3 elements contributes 6 colorings.

For elements of sort 3, there are two orbits of sizes 3 and 3 faces; there is no way to assign each orbit a single color.

For elements of sort 4, there are three orbits of sizes 2, 2, and 2 faces, so there are again 6 ways to assign the stickers so that they won't be changed by this sort of symmetry.

For the identity element, there are $\frac{6!}{2!2!2!} = 90$ distinguishable ways to put on the stickers.

The average of all these is $\frac1{24}(3\cdot6 + 6\cdot 6 + 90) = 6$, and that is the answer.

Since there are only 6, you can easily verify this by hand. The cube must have a red face, so rotate it so that this red face is on top. It has another red face, which is either on the bottom, or can be rotated to the front. Now assign the blue and yellow stickers to the remaining faces. There are 2 ways to do this if the red faces are opposite, and 4 if they are adjacent.

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