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So the example I'm trying to complete is the following:

The English alphabet contains 21 consonants and five vowels. How many strings of six lowercase letters of the English alphabet contain:

a) exactly one vowel?

My first thought was that we pick a vowel: C(5,1) Then we pick 5 out of the 21 consonants: C(21,5) or is it 21^5 (the question doesn't state that we can't re-use consonants)

Afterwards we have to sort them into a string which I think would be P(6,6)

The result I get is way too high, the answer is supposed to be (according to the answers to odd numbers) 122,523,030.

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As you say, the question doesn't say you can't reuse consonants, so it's $21^5$ for the consonants. The factor $5$ for picking a vowel is also correct. Dividing the answer you quoted by those two factors leaves a factor of $6$ unaccounted for. Can you figure out where that comes from?

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There are 5 ways ways to select exactly 1 vowel.

So For each of the 6 positions of 1 vowel there are $21^5$ words of 6 alphabets

So total words containing exactly 1 vowel = $5\cdot 6 \cdot 21^5 = 122,523,030$ (as there are 5 vowels, 6 positions of putting that vowel, and rest of the choices of consonants.)

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