We note that $ac+bd-ad-bc=a(c-d)-b(c-d)=(a-b)(c-d)$. Since each of $a, b, c, d$ is taken from the set $\{1,2,3,4,5,6,7,8,9,10\}$, then $a-b \leq 9$ since the greatest possible difference between two numbers in the set is 9 . Similarly, $c-d \leq 9$. Now, if $a-b=9$, we must have $a=10$ and $b=1$. In this case, $c$ and $d$ come from the set $\{2,3,4,5,6,7,8,9\}$ and so $c-d \leq 7$. Therefore, if $a-b=9$, we have $(a-b)(c-d) \leq 9 \cdot 7=63$. If $a-b=8$, then either $a=9$ and $b=1$, or $a=10$ and $b=2$. In both cases, we cannot have $c-d=9$ but we could have $c-d=8$ by taking the other of these two pairs with a difference of 8 . Thus, if $a-b=8$, we have $(a-b)(c-d) \leq 8 \cdot 8=64$. Finally, if $a-b \leq 7$, the original restriction $c-d \leq 9$ tells us that $(a-b)(c-d) \leq 7 \cdot 9=63$. In summary, the greatest possible value for $ac+bd-ad-bc$ is 64 which occurs, for example, when $a=9, b=1, c=10$, and $d=2$.