We can think of it in this way:
We find all cases for the equation without restriction of $A_1\leq12$ and $A_4\leq10$. Then we fill in these cases with a 2-circle Venn Diagram, the first circle ($P$) with cases satisfying $A_1>12$ (yes, $>$), and the second circle ($Q$) with cases satisfying $A_4>10$. Then what we want to find is $P'\cap Q'$ i.e. $\xi - P - Q + P\cap Q$.
Now we have to solve for
$$A_1 + A_2 + A_3 + A_4 = 25\ \fbox*$$
when $\fbox1\ A_1>12\ (A_1\geq13)$, when $\fbox2\ A_4>10\ (A_4\geq10)$, and when $\fbox3$ both occurs.
From now, we think of those 4 "$A$"s as bins, and numbers as "balls".
For $\fbox*$
Using stars and bars we get ${28\choose3}=3276$
Case $\fbox1$
Then it's easy to eliminate the restriction. We fill in box $A_1$ with 13 balls, and then we get an equation we need to solve for (with no restrictions!):
$$A_1 + A_2 + A_3 + A_4 = 12$$
Using stars and bars the answer is ${15\choose3}=455$
Case $\fbox2$
Similarly we get
$$A_1 + A_2 + A_3 + A_4 = 14$$
and hence we get ${17\choose3}=680$
Case $\fbox3$
Similarly we get
$$A_1 + A_2 + A_3 + A_4 = 1$$
and hence we get ${4\choose3}=4$
Hence we have $3276-455-680+4=2145$ which is what we want.$\ \square$