There is only one possible value for $a-c$, as you note it equals $x-y$. To count the number of solutions for $(a,b,c)$, consider the following: $a$ and $c$ are determined once we know $b$. The value of $b$ must fulfill:
- $b_\min\le b\le b_\max$ by the bounds on $b$
- $x-a_\max\le b\le x-a_\min$ by the bounds on $a$
- $y-c_\max\le b\le y-c_\min$ by the bounds on $c$
- $\lfloor \frac x2\rfloor +1\le b\le \lceil\frac y2\rceil -1$ to ensure $a.
Thus we have all in all just one constrainst $m\le b\le n$ with $m=\max\{b_\min,x-a_\max,y-c_\max,\lfloor \frac x2\rfloor +1 \}$ and $n=\min\{b_\max,x-a_\min,y-c_\min,\lceil\frac y2\rceil -1 \}$. If $n\ge m$, there are $n-m+1$ possible values for $b$, leading to the same number of solutions $(a,b,c)$. If $n, there is no solution.