What is the number of unique permutations in a list with repeated elements, if they are also considered equal under a phase shift?

Question

This is similar to many questions asked before, but none seem to need to be unique under a phase shift as well. Basically, my problem is, how many unique combinations in a group of 11 are there, if there need to be 5 A's and 6 B's. For example [A,A,B,A,B,B,B,A,B,B,A]. However, they must also have to be unique when shifted. So for example, that set would be equivalent to [A,B,A,B,B,B,A,B,B,A,A]. I hope the explanation makes sense. Thank you.

Answer 1

Imagine you place these $A$'s and $B$'s in a line of length $11$. From $11$ choose $5$ places for the $A$'s, and from the remaining $11-5=6$ choose $6$ places for the $B$'s. Because a given arrangement can be shifted over $11$ times and still be counted as just one arrangement, we divide the total number by $11$. Therefore we get $$\frac {\binom {11}{5} \binom {6}{6}}{11}=42$$ In the general case, with $n$ $A$'s and $m$ $B$'s, we have, by the same logic $$\frac {\binom {n+m}{n} \binom {n+m-n}{m}}{n+m}$$