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Earlier I was working on this question
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Earlier I was working on this question

"How many distinct anagrams are there of the word SLEEPLESSNESS?"

The answer is $\frac{13!}{2!4!5!} = 1081080$

since there are 13P13 (13!) possible permutations of 13 letters, and every one will have 4! repeats for reshuffling e's, of which each will have 5! repeats for reshuffling s's, and 2! repeats for l's, dividing through by this gives the answer.

However, how would you go about doing something like distinct three letter words? The first method works because every anagram uses every letter and as a result in the 13P13 permutations, there will always be the same about of each repeating anagram. However, for three letter words the word NPL will only be repeated twice, yet the word SSS for example, will be repeated 5P3 (60) times, and as such there is no trivial way to divide through to the answer.

How is it done, and can a general function be built? a function with 27 inputs, the occurence of each letter in the alphabet and the length of each distinct anagram?
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>>7775690
bump

the correct number for 1 letter words is of course 5, ths number of distinct letters in the word.

also for two letter words (23) i tried

$\frac{13P2}{2!2!2!}$ but it was 3 off when i tested it by writing out every distinct two letter word
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protip: just because you can put two letters together doesn't make it a word you stupid pn
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>>7775834
the point of the question is that any unique ordered triple counts as a word
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really nobody?
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well, OP here i worked them out

distinct 3 letter words: 80
distinct 7 letter words: 225,103
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I thought at least someone on this board would know what's going on.

Well, here's my working anyways, there has to be a better way than this surely
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for the seven letter distinct anagram problem
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>>7776912
bite me

>>7776905
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>>7776929
a steam game if you spot a mistake which causes one of the two answers to be wrong
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>>7776934
>apart from the one in subcase 4a i just spotted