How many 6-letter words that have either exactly 2 vowels or 4 vowels are there? (all lower case) 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!
 A: 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.
A: You have made good progress.  I will focus on case 1, which shows the issues.  When you pick two vowels, you might pick two the same.  You are correct that there are $15$ combinations of two vowels where they might be the same.  ${5 \choose 2} = 10$ of those have two distinct vowels and ${5 \choose 1}=5$ of those have two vowels the same.  If you want to compute the number of words with a specific set of four consonants, two of which are the same, you could do ${6 \choose 2}$ ways to pick the positions of the vowels, $\frac {4!}{2!}$ ways to order the consonants and $2$ or $1$ ways to order the vowels depending on whether they were the same.  The total is $12 (10 \cdot 2 + 5 \cdot 1)=300$  
I will leave this, as I think it shows a way to think about other problems.  But Eric Thoma has found a very simple solution to this one.
