# Permutation and Combinations

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.

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?
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.)