Riddle game

do i get a prize? a week in the bahamas would be very nice 

You are correct that it is not necessary to 'simplify' the string to numbering repeated characters instead of listing them.  However, that would be a little too obvious a hint.  And, if you were building a database of anagrams (i.e. useful ones not just dictionary search of random word arrangements), you would want to shorten the lookup key as much as possible for both storage savings and search performance.  Including digits from the actual original string would just require using an escape character.  For example, "November 22nd" would map to "bde2mn2ov\22" where \22 means "the character '2' two times."

You are skipping a couple steps, however, which would be quite important if you were actually coding up a solution:
1 - All letters need to be lower case, since 'n' and 'N' are not the same character but anagram searches should be case-insensitive.
2 - All spaces, punctuation, and other non-counted characters have to be removed from the string.
Both of those steps would precede the sorting, since sorting costs O(n log n) time (i.e computing time/cost scales on the order of n*log n times the length of the string, whereas lowercasing and character removal takes O(n) time (cost is linear to the length of the string).  Since removing characters shortens the string, doing the removal first saves work on the sorting.

hint: deno2rw

surely one word is an anagram of new door anyway

Nokigon wrote: By getting rid of the space?

Yes combining it to make a word...