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How many four-letter words, using the English alphabet, are possible if letters if only vowels may be repeated? How many four-letter words are there if at most one repetition of any letter is allowed? Or if exactly one repetition is allowed?

I am trying to answer these questions. I know there are $26$ letters in the english alphabet. if no letters can be repeated, it would be $26\cdot25\cdot24\cdot23$. however, my problem is if only one repetition is allowed, i don't know which letter would be repeated. $26\cdot26\cdot25\cdot24$ is different than $26\cdot25\cdot25\cdot24$. how do i get around this? I was wondering if there would be a way to subtract from the total possible number of words$(26^4)$. Or add the four letter words with no repetitions ($26\cdot25\cdot24\cdot23$) plus the ones with one repetition. Unfortunately, it seems like I’m still stuck with that one-repetition part. any help would be greatly appreciated!

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If we have exactly one repetition is allowed, we start by choosing our three letters: $\binom{26}{3}$. We then choose a letter to be repeated in $\binom{3}{1} = 3$ ways. Finally, we permute our letters in $4!/2!$ ways by the multinomial distribution. By rule of product, we multiply, to get:

$$\binom{26}{3} \cdot 3 \cdot 4!/2!$$

Or add the four letter words with no repetitions (26*25*24*23) plus the ones with one repetition.

This is correct thinking. Notice that the words with no repetition are disjoint from words with exactly one repetition. So by rule of sum, you add $26!/(26-4)!$ with the quantity I noted above.

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  • $\begingroup$ thank you so much! can you please explain the 'permute our letters' part? can we avoid that be doing P(26,3) instead of C(26,3). and why is it over 2? $\endgroup$ Commented Dec 3, 2014 at 3:57
  • $\begingroup$ You can't really avoid permuting the letters up front. At least, it's not apparent to me how it can be done. Are you familiar with the multinomial rule for counting? There are $4$ letters, and so there are $4!$ permutations. However, we have a letter appearing twice, so we divide out by $2!$. That is, $a_{1}ba_{2}b = a_{2}ba_{1}b$, where I simply label the $a$'s to illustrate the concept. $\endgroup$ Commented Dec 3, 2014 at 3:59
  • $\begingroup$ okay, i think i understand! thank you! $\endgroup$ Commented Dec 3, 2014 at 4:02
  • $\begingroup$ Glad I could help! :-) $\endgroup$ Commented Dec 3, 2014 at 4:03

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