# Number of five letter words possible using the english alphabet, not including anagrams

I'm studying for an intro to combinatorics midterm by completing past midterms. One of the questions is "How many ways are there to choose $$5$$-letter 'words' from the $$26$$-letter English alphabet with repeated letters allowed, but anagrams of other words are not allowed? For example, $$\text{TREES}$$ is an anagram of $$\text{RESET}$$, and $$\text{AAABB}$$ is an anagram of $$\text{BABAA}$$. The words need not be English language words."

I know that there are $$26^5$$ possible words that can be formed. I'm getting stuck with how to remove the total number of anagrams from this amount. Help is appreciated.

First, if all the letters of the word are different, we can choose $$5$$ letters with $$\binom{26}{5}$$ and since permutation of the letters are not counted, that's all we have.

For $$4$$ different letters, we can choose $$4$$ letters with $$\binom{26}{4}$$ and choose repeated letter with $$\binom{4}{1}$$. Therefore, we have $$\binom{26}{4}\binom{4}{1}$$ words in this case.

For $$3$$ different letters (can be chosen with $$\binom{26}{3}$$), we have two cases: Either two of the letters are repeated twice (can be chosen with $$\binom{3}{2}$$) or one of the letters is repeated three times (can be chosen with $$\binom{3}{1}$$). Therefore, we have $$\binom{26}{3}\big[\binom{3}{2}+\binom{3}{1}\big]$$ words in this case.

I think idea is clear after the above case distinction. There are two more cases for $$2$$ different letters and for only $$1$$ letter.

• When you say "all the letters are different", do you mean every letter in the English alphabet are distinct, or all of the letters in a given word are distinct? – Mr. Frothingslosh Jun 15 '19 at 21:46
• I meant $5$ letters of the word are distinct. I edited that part. – ArsenBerk Jun 15 '19 at 21:47
• We're allowed to have words with the same letter in them. AAAAA is counted as a word – Mr. Frothingslosh Jun 15 '19 at 21:48
• Yes, those words will be counted in the case for only $1$ letter mentioned at the end of answer. I didn't write the last two cases but if you have some doubts, I can add them. – ArsenBerk Jun 15 '19 at 21:49
• You're welcome :) Good luck! – ArsenBerk Jun 15 '19 at 21:58