# Math Help - Consecutive letters in anagrams

1. ## Consecutive letters in anagrams

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)

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

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

2. Originally Posted by billym
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.
Is there a cleaner/simpler way to do it?
Yes indeed there is a straightforward way.
Use the other letters as separators:^A^B^C^D^E^E^F^F^G^G^ .
We can put an H at any three of the eleven hats “^”.
And count the number of ways to rearrange the ‘separators'.
$\binom{11}{3}\cdot \frac{10!}{2^3}$