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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.

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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.

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    $\begingroup$ 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? $\endgroup$ – Mr. Frothingslosh Jun 15 '19 at 21:46
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    $\begingroup$ I meant $5$ letters of the word are distinct. I edited that part. $\endgroup$ – ArsenBerk Jun 15 '19 at 21:47
  • $\begingroup$ We're allowed to have words with the same letter in them. AAAAA is counted as a word $\endgroup$ – Mr. Frothingslosh Jun 15 '19 at 21:48
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    $\begingroup$ 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. $\endgroup$ – ArsenBerk Jun 15 '19 at 21:49
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    $\begingroup$ You're welcome :) Good luck! $\endgroup$ – ArsenBerk Jun 15 '19 at 21:58

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