I need to count the number of permutations of AABBCDEFG that have C before any instance of either B or D. My initial thought is: if it were just C before D, I could perhaps argue by symmetry that half should have this property. However, the fact that we further require C before B makes things hard in two ways. One, I'm not sure whether I could against just divide by half--something about that seems too easy. But also there are two occurrences of B and so I'm even more suspicious that this causes an extra complication.
The following argument makes intuitive sense but I wonder if it's truly valid: The fraction of permutations with the correct order of B, C, and D is the 1/(the number of arrangements of these) so equal to
$$\frac{1}{4!/2!} = 1/12$$
So that the count of all correctly ordered permutations is
$$\frac{1}{12}\cdot \frac{9!}{2!2!}$$
You are on the right track - trying to remove the extra invalid permutations by dividing.
There are $\frac{9!}{2! \cdot 2!}$ permutations of $AABBCDEFG$. Note each of these combinations has a particular ordering of $BBCD$ embedded in the combination (for each combination, remove the letters $A$, $E$, $F$, and $G$). How many of these orderings of $BBCD$ are valid? Only $3$ are - $CBBD$, $CBDB$, and $CDBB$ out of $12$ possibilities. Then you can just take $\frac{3}{12} \cdot \frac{9!}{2! \cdot 2!}$ to get the answer.