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

*

*The identity symmetry leaves fixed all $\frac{6!}{3!2!1!} = 60$ colorings of the cube.

*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.

*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.
A: First consider the three red faces.  Either one of them is opposite another, forming three-quarters of a loop, or all three are adjacent to the same vertex.  In the former case, there are three possibilities for the black face; it either "completes the loop," it's to the right of the loop's open face, or it's to the left of the loop's open face.  The two remaining faces must be white.
Edited to add:  However, as noted in the comments, the latter two possibilities are actually the same coloring, rotated a half turn around the axis through the center of the loop's open face.  Therefore, there are only two distinct colorings in this family.
If all three red faces are adjacent to the same vertex, then the two white faces will complete a loop with two of the red faces.  Use the existing coloring to choose an orientation.  The black face can be on either side of the resulting loop, resulting in two possibilities.  Edited to add:  But these are the same.  Assume the two white faces are "near" and "top" and the black face is "right."  Rotate the die a quarter-rotation toward yourself around the left-right axis (so that the two white faces are now "bottom" and "near") and then flip the die a half-rotation around the near-far axis (bringing the "bottom" face to "top" and the "right" face to the "left").
There are, therefore, five (edit: three) different colorings.
