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.
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 $$\frac{1}{24} {12\choose 2,3,3,4} + \frac{1}{4} \times 2 \times {5\choose 1, 1, 1, 2} = 11580.$$
