I think I'm finished with the anagrams except for the cases where no two consecutive letters of a given type can appear together.

For example: ABCDEEFFGGHHH

I want to find the number of arrangements that don't contain two consecutive letters H.

So: (All possible anagrams)

- (the ones with 3 consecutive Hs)

- (the ones with 2 consecutive Hs)

+ (cases of 2 consecutive Hs where the "floating" H is adjacent)

$\displaystyle \frac{13!}{2^3\cdot 3!}-\frac{11!}{2^3}-\frac{12!}{2^3}+10\cdot \frac{10!}{2^3}+4\cdot \frac{9!}{2^3}=69582240$

Assuming this is correct, is there a cleaner/simpler way to do it?