Discrete Math distinct words. "Let m, n, and r be non-negative integers. How many distinct "words" are there consisting of m occurrences of the letter A, n occurrences of the letter B, r occurrences of the letter C, such that no subword of the form CC appears, and no other letters are used?"

Step 1
Not the best possible answer, because it doesn't explain where your solution breaks down, but a different derivation...
Following the same basic principle, you want to know the number of words of the right form that have a CC in them. This has to start somewhere, and there are $n+m+r-1$ places it could start - i.e. the first C in the pair can occur as all but the last letter.
Step 2
Then you just need to fill up the remaining letters with n As, m Bs and $r-2$ Cs (so we should make some allowances for the cases $r=0$ and $r=1$, when we won't deduct anything at this stage anyway as there are no words with a CC subword).
Step 2
In the same way you found the total number of words originally, there are:
$\frac{(n+m+r-2)!}{n!m!(r-2)!}$
ways of completing the word, having put a CC somewhere. So your answer should be:
$\frac{(n+m+r)!}{n!m!r!}-\frac{(n+m+r-1)!}{n!m!(r-2)!}$
(providing $r\ge 2$).

Step 1
There are ${\displaystyle (\genfrac{}{}{0ex}{}{m+n}{m})}$ words of length $m+n$ that use m A's and n B's.
Any such word determines $m+n+1$ "gaps" (including the 2 endgaps) into which we can slip a C.
Step 2
We can choose r of these gaps in ${\displaystyle (\genfrac{}{}{0ex}{}{m+n+1}{r})}$ ways. The number of words is therefore
$(\genfrac{}{}{0ex}{}{m+n}{m})(\genfrac{}{}{0ex}{}{m+n+1}{r}).$
(By convention, ${\displaystyle (\genfrac{}{}{0ex}{}{a}{b})}=0$