<s>[INST] 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.
 [/INST] 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.
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