Hint: Imagine that we have $13$ "digits," the usual $0$ to $9$, and in addition the digits $\alpha$, $\beta$, $\gamma$, to be thought of as representing $10$, $11$, and $12$ respectively.
Pick a strictly descending sequence $x_1, x_2, x_3, x_4$ of length $4$ from this new digit set of $13$ digits. You know how to count how many such strictly descending sequences there are.
Let $a_4=x_4$, $a_3=x_3-1$, $a_2=x_2-2$, $a_1=x_1-3$. Show that this produces every non-increasing sequence $a_1\ge a_2 \ge a_3 \ge a_4$ in one and only one way.
Now we need to adjust for the unpleasant fact that the initial digit of a $4$-digit number cannot be $0$. That translates into the fact that the strictly descending sequence $x_1=3$, $x_2=2$, $x_3=1$, $x_4=0$ is not allowed.
Remark: The above procedure has nothing much to do with digits. The number of never increasing sequences of length $k$ from the set $\{1,2,\dots,n\}$ is the same as the number of strictly decreasing sequences of length $k$ from the set $\{1,2,\dots,n,n+1,\dots, n+k-1\}$.
Added: The number of strictly decreasing sequences of length $4$ from the digits $0, 1,\dots, 9$ is just $\binom{10}{4}$, the number of ways to choose $4$ distinct numbers. If we use the "digits" $0,1,\dots,9,\alpha,\beta,\gamma$, then the number of strictly decreasing sequences is $\binom{12}{4}$. We must subtract $1$ because $3,2,1,0$ is not allowed.
How many integers solutions of $$x_1 + x_2 + x_3 + x_4 = 28$$ are there that satisfy $-10 \leq x_k \leq 20$ for $1 \leq k \leq 4$?
We can convert this to an equivalent problem in the nonnegative integers.
$$x_1 + x_2 + x_3 + x_4 = 28 \tag{1}$$
Let $y_k = x_k + 10$, where $1 \leq k \leq 4$. Then each $y_k$ is a nonnegative integer satisfying $0 \leq k \leq 30$. Substituting $y_k - 10$ for $x_k$, $1 \leq k \leq 4$, in equation 1 yields
\begin{align*}
y_1 - 10 + y_2 - 10 + y_3 - 10 + y_4 - 10 & = 28\\
y_1 + y_2 + y_3 + y_4 & = 68 \tag{2}
\end{align*}
Equation 2 is an equation in the nonnegative integers. A particular solution of equation 2 corresponds to the placement of $3$ addition signs in a row of $68$ ones. The number of solutions of equation 1 is
$$\binom{68 + 3}{3} = \binom{71}{3}$$
since we must choose which $3$ of the $71$ positions required for $68$ ones and $3$ addition signs will be filled with addition signs.
From these, we must subtract those cases that violate one or more of the restrictions that $y_k \leq 30$, $1 \leq k \leq 4$. Notice that at most two of these restrictions can be violated simultaneously since $3 \cdot 31 = 93 > 68$.
Suppose $y_1 > 30$. Then $y_1' = y_1 - 31$ is a nonnegative integer. Substituting $y_1' + 31$ for $y_1$ in equation 2 yields
\begin{align*}
y_1' + 31 + y_2 + y_3 + y_4 & = 68\\
y_1' + y_2 + y_3 + y_4 & = 37 \tag{3}
\end{align*}
Equation 3 is an equation in the nonnegative integers with
$$\binom{37 + 3}{3} = \binom{40}{3}$$
solutions. By symmetry, there are an equal number of solutions for each of the four variables that could violate the restriction $y_k \leq 30$. Hence, there are
$$\binom{4}{1}\binom{40}{3}$$
cases in which one of the restrictions is violated.
However, if we subtract this number from the total, we will have subtracted too much since we will have subtracted each case in which two of the variables violate the restrictions twice, once for each variable we designated as the variable that violated the restriction. We only want to subtract those cases once, so we must add them back.
Suppose $y_1, y_2 > 30$. Let $y_1' = y_1 - 31$; let $y_2 = y_2 - 31$. Then $y_1'$ and $y_2'$ are nonnegative integers. Substituting $y_1' + 31$ for $y_1$ and $y_2' + 31$ for $y_2$ in equation 2 yields
\begin{align*}
y_1' + 31 + y_2' + 32 + y_3 + y_4 & = 68\\
y_1' + y_2' + y_3 + y_4 & = 6 \tag{4}
\end{align*}
Equation 4 is an equation in the nonnegative integers with
$$\binom{6 + 3}{3} = \binom{9}{3}$$
solutions. By symmetry, there are an equal number of solutions for each of the $\binom{4}{2}$ pairs of variables that could violate the restrictions. Hence, there are
$$\binom{4}{2}\binom{9}{3}$$
cases in which two of the variables violate the restrictions.
Thus, by the Inclusion-Exclusion Principle, the number of solutions that satisfy the restrictions is
$$\binom{71}{3} - \binom{4}{1}\binom{40}{3} + \binom{4}{2}\binom{9}{3}$$
Best Answer
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.$\ \square$