How many $3$-letter words are there with no repeated letter if the middle letter is a vowel

combinatoricsdiscrete mathematics

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! 🙂

Best Answer

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 $5 \times 21 \times 20$.

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