Find the number of different colorings of the faces of the cube with 2 white, 1 black, 3 red faces?

Question

I tried using Polya theorem same in this https://nosarthur.github.io/math%20and%20physics/2017/08/10/polya-enumeration.html guide, using S4 group for cube faces. But i Have a $\frac{1}{24}12w^2*b*r^3$. Please help to me how count coloring faces of the cube and other platonic solids.

Answer 1

The cycle index polynomial for rotational symmetries of the faces of a cube is $$ \tfrac1{24}(z_1^6+12z_2^3+8z_3^2+3z_1^2z_2^2) $$ Therefore, in order to count the number of colorings with one black, three red, and two white faces, we need to find the coefficient $b^1r^3w^2$ in $$ \tfrac1{24}[(b+w+r)^6+12(b^2+r^2+w^2)^3+8(b^3+r^3+w^3)^2+3(b+r+w)^2(b^2+r^2+w^2)^2] $$ Going term by term...

• The coeficient of $b^1r^3w^2$ in $(b+r+w)^6$ is given by the multinomial coefficient $\binom{6!}{1!3!2!}=60$.

• $(b^2+r^2+w^2)^3$ contributes nothing to $b^1r^3w^2$.

• $(b^3+r^3+w^3)^2$ also contributes nothing.

• The trickiest term is $(b+r+w)^2(b^2+r^2+w^2)^2$. I will write this as $$ (b+r+w)(b+r+w)(b^2+r^2+w^2)(b^2+r^2+w^2) $$ In order to contribute to $b^1r^3w^2$, the $(b^2+r^2+w^2)$ factors need to contribute an $r^2$ and a $w^2$. This can be done in either order, for $2$ choices. The remaining $b^1r^1$ comes from the $(b+r+w)$ factors; again, either one can give the $b$ while the other gives an $r$, for another $2$ choices. Therefore, the contribution from this term is $2\times 2=4$.

All in all, the coefficient is $$ \frac1{24}[1\cdot 60+12\cdot 0+8\cdot 0+3\cdot 4]=\boxed{3} $$

Here is a Wolfram Alpha computation which confirms this.

Answer 2

For this we can use the Burnside lemma, a simpler version of the Pólya theorem. We consider the five conjugacy classes of symmetries of the cube and for each symmetry we count the number of colorings o the cube left fixed by the symmetry. Consider the black face. For a coloring to be left fixed by some symmetry, the black face must be left fixed. This rules out the rotations where the axis goes through a vertex or an edge of the cube, since those fix no faces. So we need only consider:

1. The identity symmetry leaves fixed all $\frac{6!}{3!2!1!} = 60$ colorings of the cube.
2. There are 6 rotations by 90° around an axis through a face of the cube. These divide the faces into orbits of sizes $4, 1, 1$. There is no way for such a rotation to leave the three red faces fixed.
3. There are 3 rotations by 180° around an axis through a face of the cube. These divide the cube into orbits of sizes $2, 2, 1, 1$. For one of these rotations to leave the coloring unchanged, The black face must be in one of the two orbits of size $1$ and the white faces in one of the two orbits of size $2$. There are $2·2=4$ ways to do this and then the red faces are in the other two orbits.

The total is $$\frac1{24}\left(\underbrace{60}_{\text{identity}} + \underbrace{6·0}_{90°} + \underbrace{3·4}_{180°}\right) = 3.$$

Note that this calculation is completely analogous to the one done by Mike Earnest elsewhere in the thread.