Put separate stickers on each of the four 'A', three 'B', and one of the two C's, to distinguish them.
Now there are 9 distinct letters can be arranged in 9! ways. Remove the stickers, of all A's, then all B's and then all C's. Each time we remove the stickers, 4! arrangements collapse into 1 for A's, 3! arrangement collapses to 1 for B's and 2! to 1 for C.
So the number of arrangements the given 9 letters is \[ \large \frac{9!}{4! \times 3! \times 2!} \]