Coloring the windmill A windmill has $5$ wings and each of these is symmetrically connected to the axis and consists of two parts.
If on the wings of the windmill $4$ parts are colored black, $3$ parts are colored red, and $3$ parts are colored orange, in how many ways can one then color the windmill? (In different ways when one looks at the windmill from the front and the wings rotate.)
                                    
Use Pólya's theorem and the fact that the group $C_n$ that consists of rotations of a regular $n$-polygon has the cycle index
$$\frac 1n\sum_{d\,|\, n} \varphi(d)t_d^{\frac nd},$$
where $\varphi(d)=\left |\{j\,:\, 0\leq j\leq d-1,\; \textrm{gcd}(j,d)=1\}\right |$, but observe that in this case the number of orbits is double that of a regular $n$-polygon because each wing is built from two parts.
Here $\sum_{ d\,|\,n}$ means that we take the sum over all integers $d$ such that $1\leq d\leq n$ and $n$ is divisible by $d$ and the term $t_d^{\frac nd}$ is due to the fact that in the rotation group there is an element so that the cyclic group generated by it has $\frac nd$ orbits with length $d$.
Pólya's theorem says that if in the cycle index the variable $t_j$ is replaced by $b^j+r^j+o^j$ then the coefficient of the term $b^4\cdot r^3\cdot o^3$ is the number one is looking for!
My attempt:
I set $n = 4 + 3 + 3 = 10$ and determined all divisors $d$ of $n$ to be $1,2,5,10$. Now $\varphi (1) = 1,\,\varphi (2) = 1,\,\varphi (5) = 4,\,\varphi (10) = 4$. Plugging into equation i got the cycle index $$\frac{1}{{10}}\left( {t_1^{10} + t_2^5 + 4t_5^2 + 4{t_{10}}} \right)$$
Calculating for $b^4\cdot r^3\cdot o^3$ 
$$\frac{1}{{10}}\left( {{{(b + r + o)}^{10}} + {{({b^2} + {r^2} + {o^2})}^5} + 4{{({b^5} + {r^5} + {o^5})}^2} + 4({b^{10}} + {r^{10}} + {o^{10}})} \right)$$
I get the coefficient for $b^4\cdot r^3\cdot o^3$ $420$, which is wrong. Now, could someone explain my mistake?
 A: You have an action of $\mathbb Z_5$ on the set of all colorings of the windmill (a set with $\binom{10}{4,3,3}$ elements).
By Burnside the number of orbits is:
$$\frac{1}{|G|}\sum\limits_{g\in G}X^g.$$
There are only two types of permutations in play.
The identity clearly leaves all $\binom{10}{3,3,4}$ colorings unchanged.
Any other rotation:
This permutation leaves a coloring fixed if and only if all of the wings are equal, and this is impossible, so it leaves $0$ colorings unchanged.
Therefore the number of orbits is $\frac{1}{5}\cdot\binom{10}{3,3,4}=\frac{10!}{3!3!4!5}=840$.

In fact we don't really need Burnside here. There are $\binom{10}{4,3,3}$ ways to color the windmill if it didn't rotate. Every "movable" windmill gives us $5$ different positions. Therefore there are $\binom{10}{3,3,4}/5$ "movable" windmills.
A: Let me typeset some of the content from the comments. We have five panels attached to the center and five is prime, so the cycle index is really simple here namely $$Z(C_5) = \frac{1}{5} (a_1^5 + 4 a_5).$$ Taking into account that the outer and inner elements of the panels move in sync but never enter into the same orbit we get $$Z(W) = \frac{1}{5} (a_1^{10} + 4 a_5^2).$$
Now we are looking for $$[B^4 R^3 O^3] Z(W)(B+R+O).$$
We get for $$Z(W)(B+R+O) = 
1/5\, \left( B+R+O \right) ^{10}+4/5\, \left( {B}^{5}+{O}^{5}+{R}^{5}
 \right) ^{2}$$
and $$[B^4 R^3 O^3] Z(W)(B+R+O) = 840.$$
Note that the second term produces multiples of five in the exponents and hence does not contribute to the coefficient, which is
$$\frac{1}{5} {10\choose 4,3,3} = 840.$$
