I am not sure about circular permutations, but this problem seems similar to the well-known question about the number of anagrams (i.e., all permutations) of a word. For example, "mathematics" has 11 letters, so if all of them were different, there would be 11! permutations. However, it has 2 a's, 2 t's and 2 m's. Therefore, those 11! permutations have to be broken into equivalence classes. If two words belong to the same class, they may differ by the order of a's, t's and m's -- they are still the same word because we don't distinguish two a's. Each class has 2!*2!*2! words, so the number of classes is 11!/(2!*2!*2!).