Pascal's Triangle is awesome for solving pathing problems like this.

To solve this, the easiest way is to use Pascal's Triangle as mentioned in the hint. It works well because each t(n,r) cell value represents the number of ways to reach that cell.

First, I used the grid provided to structure the frame of an upside-down Pascal's Triangle (i.e. filling in the ones on the outer layer). In the same step, I filled in the first few rows of the triangle because they follow the same summative t(n,r)=t(n-1,r-1)+t(n-1,r) pattern.

Past the third row (technically fourth because B counts as 1), the pattern stops working on the edges because there's no second value to add. To resolve this, we can always "add" zero, making it the same as the previous term.

Again, our pattern starts to break down but this time because of the white pieces. This may seem like a difficult edge case, but we can simply use the diagonal preceding the white piece, provided that isn't also a white piece (which is why the top left square stays as zero).

Finally, to get the answer we simply sum the first row.