Show that a regular hexagon’s edges may be coloured red, white or blue in $92$ essentially different ways. How many ways are possible if an equal number of red, white and blue edges must appear?
I have done the first part and got $92$ using Burnside's lemma. But I'm not sure about the second part - with the added constraint of exactly $2$ edges of each colour, it seems much more difficult to count how many such colourings are fixed under each symmetry.
There are $\binom6{2,2,2}=90$ different colourings. The identity fixes all $90$ of them. A rotation of order $6$ has a single orbit, and a rotation of order $3$ only has two orbits, so neither fixes any colourings. A rotation of order $2$ (of which there is $\phi(2)=1$) has three orbits of size $2$, and we can assign the three colours to these three orbits in $3!=6$ ways.
A reflection in a line through opposite midpoints (of which there are $\frac62=3$) also has three orbits of size $2$, and thus also fixes $6$ colourings. A reflection in a line through opposite vertices (of which there are also $\frac62=3$) has two orbits of size $2$ and two fixpoints. Since two of the colours have to be used for the orbits of size $2$, the two fixpoints have to be the third colour; so this also yields $3!=6$ invariant colourings.
In total, that makes
$$ 90+6+3\cdot6+3\cdot6=132\;, $$
and dividing by the size of the group, $12$, yields a count of $11$ equivalence classes of colourings.
Okay, I think I finally get it. So, you have dihedral group $D_{12}$ and you're counting $3-$collourings of hexagon, which are different with respect to your group. For the case of using any colour you like, then you should get something like $\frac{1}{12}(3^6 + 2\cdot 3 + 2\cdot 3^2 + 3^3 + 3 \cdot 3^2 \cdot 3^2 + 3 \cdot 3^3)$. Before giving value $3$ to every cycle you meet during analysing that hexagon, you should get a polynomial which looks like:
$P(x_1,x_2,x_3,x_4,x_5,x_6)=\frac{1}{12}(x_1^6 + 2x_6 + 2x_3^2 + x_2^3 + 3x_1^2x_2^2 + 3x_2^3)$ which is called cycle index.
monomial $x_k$ occurs when the cycle in your permutation has length $k$ and you're raising it to the power which is equal to number of those cycles in your permutation. Now, you want to find such collouring that there are $2-2-2$ proportions. We will analyse the number of cycles and their lengths one by one.
There are ${6 \choose 2}{4 \choose 2}$ possibilities for $x_1^6$ terms
There are no possibilities for $x_6$ term
There are no possibilities for $x_3^2$ term
There are $3!$ possibilities for $x_2^3$ term
There are $3!$ possibilities for $x_1^2x_2^2$ term
And finally there are $3!$ possibilities for $x_2^3$ term.
Which gives $\frac{1}{12}({6 \choose 2}{4 \choose 2} + 6 + 18 + 18) = \frac{132}{12} = 11$ different $2-2-2$ collourings.
To explain a little. When we counted the possibilities for terms, it was crucial that we needed to assign one colour for each cycle (so we need at least $3$ cycles to have a collouring). So in fact we were collouring cycles. And for example with term $x_2^3$ we have $3$ of length $2$, so we needed to just choose which colour assign to which cycle, what can be done $3\cdot 2\cdot 1$ ways, and for example in term $x_1^2x_2^2$ we needed to choose colour for two cycles of length $1$ (note that $1$ colour need to have both of those cycles), which could be done in $3$ ways, and then for the rest $2 \cdot 1$ ways.
You can google Polya-enumerative theorem or cycle index. There are many more thing you can do with it.
