Presumably, you don't want to count substrings that would be formed by the removal of the Xs?

Also, you mentioned earlier having all the permutations, including those with zero occurances, in the results hash. The problem with that is that by the time you get to n=13, you'd have 67 million entries in the hash, most of which would be 0.

Anyway, this might be of some use:

$s = 'aaaabccbaxxxcbddadcc';;

++$h{ $1 } while $s =~ m[(?=([abcd]{3})).]g;;

pp \%h;;
{
  aaa => 2,
  aab => 1,
  abc => 1,
  adc => 1,
  bcc => 1,
  bdd => 1,
  cba => 1,
  cbd => 1,
  ccb => 1,
  dad => 1,
  dcc => 1,
  dda => 1,
}
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