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Math Help - Consecutive letters in anagrams

  1. #1
    Member billym's Avatar
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    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?
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  2. #2
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    Quote Originally Posted by billym View Post
    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}
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