Number of bracelets of length $12$ that can be obtained from “AABBBCCCDDDD” by permutation A similar question about necklaces has already been answered. I would like to know the very same thing regarding bracelets; that is, letting strings be equivalent under reflection.
For my practical purpose, I need to know the number of bracelets of length $12$ that can be obtained from “AABBBCCCDDDD” by permutation.
 A: There is quite a number of these at
MSE meta
written by  different users using  several equivalent variants  of the
standard notation.
Basically  you need  to  compute  the  cycle index  of  the
dihedral  group   on  twelve  elements  $Z(D_{12})$   and  do  careful
coefficient extraction. The answer is
$$[A^2 B^3 C^3 D^4] Z(D_{12})(A+B+C+D).$$
The cycle index is
$$Z(D_{12}) =
1/24\,{a_{{1}}}^{12}+{\frac {7\,{a_{{2}}}^{6}}{24}}+1/12\,{a_{{3}}}^
{4}+1/12\,{a_{{4}}}^{3}\\+1/12\,{a_{{6}}}^{2}+1/6\,a_{{12}}+1/4\,{a_{{
1}}}^{2}{a_{{2}}}^{5}$$
Recall the usual cycle index substitution which in the present case is
$$a_q^p = (A^q + B^q + C^q + D^q)^p.$$
Now $a_3^4$  produces multiples of three  in the exponents  so it does
not contribute. The terms in $a_2^6, a_4^3, a_6^2$ and $a_{12}$ do not
either  for  similar  reasons  (we  have  $\gcd(2,3,3,4)  =  1.$)  The
contribution from $a_1^{12}$ is
$$\frac{1}{24} {12\choose 2,3,3,4}.$$
The  term $a_1^2  a_2^5$ contributes  as follows.  As we  have  an odd
number of Bs and Cs one of each must come from $a_1^2$ which yields
$$\frac{1}{4} \times 2 \times {5\choose 1, 1, 1, 2}.$$
Adding these two yields the answer
$$\frac{1}{24} {12\choose 2,3,3,4}
+ \frac{1}{4} \times 2 \times {5\choose 1, 1, 1, 2}
= 11580.$$
