Burnsides Theorem Application - Necklace Having a lot of trouble understanding this question: How many different necklaces can be made from eight black beads and four white beads?
What I know:
Burnsides Theorem is given by: $$\frac{1}{|G|} \sum_{g \in G} |\operatorname{fix}_{\Omega} (g)|$$
8+4=12 and this corresponds to the Dihedral group of order 12 (I use the notation $D_{2n})$ so I will note this as $D_{24}$, also known as the symmetric 12-gon. 
So we know that $|G|=24$ and the total number of necklace configurations is given by $\binom{12}{4}=495$ so I'm guessing we have to $Fix(e)=495$? But I'm unsure how to apply the rest of Burnsides theorem from here and work out the configurations of rotations and reflections. 
Any help would be great, thanks.

Given that  we have dihedral symmetry  we suppose we are  working with
bracelets (naming convention by OEIS).   This requires the cycle index
$Z(D_{12})$ of  the dihedral  group $D_{12}.$ We  have for  the cyclic
group that
$$Z(C_{12}) = \sum_{d|12} \varphi(d) a_d^{12/d}
= \frac{1}{12}
(a_1^{12} + a_2^6 + 2 a_3^4 + 2 a_4^3 + 2 a_6^2 + 4  a_{12}).$$
We get twelve  more permutations corresponding to flips  about an axis
passing through opposite slots or opposite edges, for a result of
$$Z(D_{12}) = \frac{1}{24}
(a_1^{12} + a_2^6 + 2 a_3^4 + 2 a_4^3 + 2 a_6^2 + 4  a_{12})
+ \frac{1}{24} ( 6 a_1^2 a_2^5 + 6 a_2^6).$$
By  the Polya  Enumeration  Theorem  (PET) we  are  interested in  the
quantity
$$[B^8 W^4] Z(D_{12}; B+W).$$
Working through the terms we find

 $[B^8 W^4] (B+W)^{12} = {12\choose 8},$ 
 $[B^8 W^4] (B^2+W^2)^{6} = [B^4 W^2] (B+W)^{6}
=  {6\choose 4},$ 
 $[B^8 W^4] 2 (B^3+W^3)^{4} = 0$ 
 $[B^8 W^4] 2 (B^4+W^4)^{3} = [B^2 W] 2 (B+W)^{3}
= 2 {3\choose 2}$ 
 $[B^8 W^4] 2 (B^6+W^6)^{2} = 0$ 
 $[B^8 W^4] 4 (B^{12}+W^{12}) = 0$ 

We get from the reflections

 $[B^8 W^4] 6 (B+W)^2 (B^2+W^2)^5
\\ = [B^6 W^4] 6 (B^2+W^2)^5 + [B^7 W^3] 12 (B^2+W^2)^5 +
[B^8 W^2] 6 (B^2+W^2)^5
\\ = [B^3 W^2] 6 (B+W)^5 + [B^4 W] 6 (B+W)^5
\\ = 6 {5\choose 3} + 6 {5\choose 4}$ 
 $[B^8 W^4] 6 (B^2+W^2)^{6} = [B^4 W^2] 6 (B+W)^{6}
=  6 {6\choose 4}.$ 

Collecting everything we find for our result that it is
$$\frac{1}{24} \left({12\choose 8} + {6\choose 4} + 2 {3\choose 2}
+ 6{5\choose 3} + 6{5\choose 4} + 6 {6\choose 4}\right).$$
which yields
$$\bbox[5px,border:2px solid #00A000]{29.}$$
