Hey there!
When we make a selection of letters then look at combinations, it's important to note that we need to choose the letters we arrange first, before arranging them to form combinations. The issue with this question is there are duplicate A's, so we need to consider cases in terms of how many A's we get. What does make it easier is that though there are four A's, there are only ever three letters chosen, which reduces the number of cases.
First note that there is ever only one way to pick any number of A's, since they're all the same. We just need to pick the 6 other filler letters depending on how many A's we have out of three ie. if there is one A, we need two filler letters.
Thus, the number of ways will just be the number of ways to pick the filler letters, then rearrange the whole set itself - 6C3 x 3! + 6C2 x 3! + 6C1 x 3 + 6C0 x 1.
Hope this helps