<s>[INST] How to color a cube with exactly 6 colors using Polya enumeration theory I have seen in other questions (Painting the faces of a cube with distinct colours) , and found for myself there are 30 colorings of the cube with exactly 6 colors. My issue is when I use Polya enumeration theory I take the number of colorings up to 6 to be $\frac{1}{24}(6^6+3*6^4+12*6^3+8*6^2)$ (https://en.wikipedia.org/wiki/P%C3%B3lya_enumeration_theorem) and subtract the number of colorings up to 5, $\frac{1}{24}(5^6+3*5^4+12*5^3+8*5^2)$ to get 1426. What is wrong with my application of Polya enumeration theory? How could I fix it?
 [/INST] To find the number of ways to color the faces of a cube with exactly $6$ colors using Polya's Enumeration Theorem:
First, as you already know, the cycle index of the symmetries of the face permutations of a cube is
$$Z = \frac{1}{24}(x_1^6 + 6x_1^2x_4+3x_1^2x_2^2 + 8x_3^2 + 6x_2^3)$$
The figure inventory for $6$ colors represented by $a,b,c,d,e,f$ is
$$a+b+c+d+e+f$$
"Substituting" the figure inventory into the cycle index, we have
$$\frac{1}{24} \left((a+b+c+d+e+f)^6 \\ \\+6 (a+b+c+d+e+f)^2
   \left(a^4+b^4+c^4+d^4+e^4+f^4\right) \\+3 (a+b+c+d+e+f)^2 
   \left(a^2+b^2+c^2+d^2+e^2+f^2\right)^2\\+8
   \left(a^3+b^3+c^3+d^3+e^3+f^3\right)^2 \\ +6 \left(a^2+b^2+c^2+d^2+e^2+f^2\right)^3  \right) $$
The number of ways to color the cube using each of the six colors exactly once is the coefficient of $abcdef$ when this polynomial is expanded.  In difficult cases I would use a computer algebra to find the coefficient, but in this example it's easy to see that what we want is $1/24$ the coefficient of $abcdef$ in $(a+b+c+d+e+f)^6$, i.e.
$$\frac{6!}{24} = 30$$
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