Build prefix sums on the frequency of Q. You can then calculate the number of subsequences by summing up number of prefix Q's multiplied by the number of suffix Q's for each position where A is present.

There is an editorial about this.

If we only iterate on the place of 'A', we can get the number of 'Q' before and after it using prefix sums, and it leads to O(n) solution.

If you dont know how to implement, here it is

int prefix = 0;                  /// Number of 'Q' on the left  side
int suffix = count(all(s), 'Q'); /// Number of 'Q' at the right side
int res = 0;
for (char c : s)
{
    if (c == 'Q')
    {
         ++prefix;
         --suffix;
    }
    else /// c = 'A'
        res += prefix * suffix; /// Number of ways to select left * right.
}