What is the number of essentially different ways to paint the faces of a cube such that one face is red, two are blue, and the remaining three are green?
Do this ‘by hand’ (which may be quicker) as well as by using the cycle index theorem, and compare your answers.
Using Burnside's Lemma, I calculated the cycle index for the faces of a cube and that turned out to be \begin{align*} Z(G) = \frac{1}{24}(a_1^6+3a_1^2a_2^2+6a_1^2a_4+6a_2^3+8a_3^2) \end{align*} and with $n=3$, there are $57$ different ways to color the cube, but I'm not sure how to go from here.
Any help, tips, or a fully worked out solution would be appreciated!
Yes, we may use the multivariate generating function $$ Z_G(a_1,a_2,a_3,a_4) = \frac{1}{24}(a_1^6+3a_1^2a_2^2+6a_1^2a_4+6a_2^3+8a_3^2)$$ by letting $a_i=r^i+b^i+ g^i$ for $i=1,\dots,4$. Then we extract the coefficient of $r^1b^2g^3$ which is just the number of ways, up to rotations, to paint the faces of the cube such that $1$ face is red, $2$ are blue, and $3$ are green. $$[r^1b^2g^3]Z_G=[r^1b^2g^3]\frac{a_1^6+3a_1^2a_2^2}{24}=[b^2g^3]\frac{6(g+b)^5+6(g+b)2(g^2b^2)}{24} =\frac{6\binom{5}{2}+12}{24}=3.$$
Further explanation. We paint one face red and then we have two cases:
1) If the opposite side is blue then we can paint the remaining blue face in $1$ way: it is one of the four lateral faces.
2) If the opposite side is green then we can paint the four lateral faces in $2$ ways: alternating blue and green faces or two adjacent blue faces and two adjacent green faces.
The total number of ways is $1+2=3$.