My first answer was $5 \times 25 \times 24$, picking the vowel before the first and last letter. To my surprise, my book says this is the right answer! But I thought I was not counting everything and continued considering the following situations:

  1. If I pick one vowel, there are $5 \times 21 \times 20$ words since I'm not considering vowels in the remaining choices. For instance, suppose I pick "abd". Then, there are $3!$ ways to arrange this word, but I want the middle letter to be the vowel, so only "bad" and "dab" are valid. Then for each of the $5 \times 21 \times 20$ words, there are two valid permutations. Therefore, there are $5 \times 21 \times 20 \times 2$ words.

  2. If I pick two vowels, there are $5 \times 4 \times 21$ words. For instance, suppose I pick "aed". Again, there are $3!$ ways to arrange this word, but since there are two vowels, we have four valid permutations: "aed", "dea", "ead", and "dae". So for each of the $5 \times 4 \times 21$ words, there are four valid permutations. Therefore, there are $5 \times 4 \times 21 \times 4$ words.

  3. If I pick three vowels, there are $5 \times 4 \times 3$ words. Since all are vowels, all $3!$ permutations are valid. Therefore, there are $5 \times 4 \times 3 \times 6$ words.

Finally, I summed all three to get $(5 \times 21 \times 20 \times 2) + (5 \times 4 \times 21 \times 4) + (5 \times 4 \times 3 \times 6)$ $3$-letter words with no repeated letters such that the middle letter is a vowel. If the answer in the book is correct, clearly I overcomplicated a simple problem, but I cannot see why the book is correct. Am I overcounting? Am I considering cases that should not be considered?

Thank you for any clarifications! :)

  • 1
    $\begingroup$ Remember, that the important vowel must be in the middle. Since this one is not free to be in any other position, its choice is completely independent of the first and last letter choices. $\endgroup$ Sep 15, 2020 at 0:21
  • $\begingroup$ What does "repeated" mean here? Is "PUP" allowed? $\endgroup$ Sep 15, 2020 at 0:46
  • $\begingroup$ "PUP" would not be allowed because you repeated "P" two times. $\endgroup$ Sep 15, 2020 at 1:10

1 Answer 1


The numbers in your approach are not correct: if we really want to break things down into cases by number of vowels, we should get $$ (5 \times 21 \times 20) + (5 \times 4 \times 21\times 2) + (5 \times 4 \times 3) $$ and this simplifies to $3000$, same as the textbook answer of $5 \times 25 \times 24$.

Your mistake is mainly multiplying by permutations in cases where that's already accounted for. For example, in the one-vowel case: there are $5$ ways to pick the vowel (which must go in the middle), then $21$ ways to pick the consonant at the beginning, then $20$ ways to pick the consonant at the end. This already specifies the order the letters go in. You don't have to multiply by $2$ to account for symmetry.

Or, if you prefer: we could pick the vowel in $5$ ways, then pick two consonants in $\binom{21}{2}$ ways, then order the consonants in $2!$ ways, to give $5 \times \binom{21}{2} \times 2$ for this case. This is equal to $5 \times 21 \times 20$.

  • $\begingroup$ Thank you. Your last paragraph was very enlightening. If I generalize this approach for the cases where I pick two and three vowels, I would have ${5 \choose 2} \times 21 \times 2!^2$ and ${5 \choose 3} \times 2!^3$ respectively, right? $\endgroup$ Sep 15, 2020 at 1:09
  • $\begingroup$ The last one should be $3!$, not $2!^3$: any permutation of the $3$ vowels is good. $\endgroup$ Sep 15, 2020 at 1:17
  • $\begingroup$ Oh that's right, silly mistake. So each case is different, it seems there is no general rule... Well, the first answer $5 \times 25 \times 24$ is indeed the simplest to understand. Thank you for your time! :) $\endgroup$ Sep 15, 2020 at 1:24

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