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
$$\bbox[5px,border:2px solid #00A000]{29.}$$
