John walks from point $A$ to $C$ while Mary goes from point $B$ to $D$. Both of them will move along the grid, either right or up, so they take shortest routes. How many different possibilities are there such that their routes do not intersect?
This problem can be solved in many different ways. One of them is to count the opposite cases using bijection.
Without any restriction, there are $\binom{8}{4}$ different paths for John and the same number of paths for Mary. Hence, there are total $\binom{8}{4}^2$ different combinations.
Next, instead of counting the number of paths that do not intersect, let's count the number of intersecting paths. Assuming two such paths first insect at point $X$. Then the original paths $(AXC,\ BXD)$ will have a bijection with $(AXD,\ BXC)$. (Imagine that John and Mary switch their destinations at the first intersection point.)
Because there are $\binom{10}{4}$ paths from $A$ to $D$, and $\binom{6}{2}$ ways from $B$ to $C$, so the number of intersecting paths is $\binom{10}{4}\binom{6}{2}$.
It follows that the final answer is $$\binom{8}{4}^2 - \binom{10}{4}\binom{6}{2}=\boxed{1750}$$