I considered two cases.
Case 1 (2 vowels): Pick 2 vowels $\binom{5 + 2 -1}{2}$, then pick 4 consonants $\binom{21 + 4 -1}{4}$, then order them $6!$.
Case 2 (4 vowels): Pick 4 vowels $\binom{5 + 4 -1}{4}$, then pick 2 consonants $\binom{21 + 2 -1}{2}$, then order them $6!$.
Total: $\binom{5 + 2 -1}{2} \binom{21 + 4 -1}{4} 6! +\binom{5 + 4 -1}{4} \binom{21 + 2 -1}{2} 6!$
But then I realized I couldn't do the factorial step because I could have the same vowels/consonants appearing more than once so I would be over counting their order. How can I clear this up?
Is there an alternative way I can go about this?
Thanks!
Best Answer
We can get around it as follows for the two vowel case:
Represent the word as six blank spots: _ _ _ _ _ _.
Choose two spots for the vowels: ${6 \choose 2}$.
Since order matters, we can fill those two spots in $5^2$ ways.
Since order matters, we can fill the remaining consonants in $21^4$ ways.
For a total of ${6 \choose 2} 5^2 21^4=72930375$ ways.
A similar argument applies to the four vowel case. I get a final total of $77064750$ different words with either exactly two or exactly four vowels.