Each body is allowed a jump of $+1 \uparrow$ or $+1 \rightarrow$ in one step. There are $n$ steps available to each body. Find the number of non-intersecting path pairs between a fixed source and a fixed destination within this grid.
Assume final destination is $X,Y$ and of course, $X + Y = n$.
My approach
Let $\Delta X_1, \Delta Y_1$ and $\Delta X_2, \Delta Y_2$ be the displacement vectors for body 1 and body 2 respectively.
Let $Array(\Delta X_1)$ denote the array containing the values of $\Delta X_1$ indexed by step, and so on for other vectors too. Note that there is a bijection between $Array(\Delta X_1)$ and $Array(\Delta Y_1)$, because a body can jump only along one axis at a time.
So we have to consider only $Array(\Delta X_1)$ and $Array(\Delta X_2)$ to find the number of paths. The paths intersect if at any particular step, $\Delta X_1 = \Delta X_2$, which also implies $\Delta Y_1 = \Delta Y_2$ from the bijection (so ideally, this should happen only at the source and the destination). $Array(\Delta X_1)$ is $n$ sized and monotonic increasing over its length and $Set(Array(\Delta X_1))$ is $\mathbb{N}_{\leq X}$.
An example of a pair of non-intersecting paths for source $(0,0)$ and destination $(4,4)$ is-
Body 1 - $Array(\Delta X_1) = \{ 0,0,0,1,1,2,3,4\}$ and $Array(\Delta Y_1) = \{ 1,2,3,3,4,4,4,4\}$
Body 2 - $Array(\Delta X_2) = \{ 1,1,2,3,3,4,4,4\}$ and $Array(\Delta Y_2) = \{ 0,1,1,1,2,2,3,4\}$
This scenario can be condensed to a series of $\rightarrow$ and $0$ (we need to consider only one of the $x$ and $y$ displacement vectors, as the latter can be derived uniquely from the former). If the displacement is $+1$, place a $\rightarrow$ on the final coordinate, else $0$.
So - Body 1 - $Array(\Delta X_1) = \{ 0,0,0,\rightarrow,0,\rightarrow,\rightarrow,\rightarrow\}$ and Body 2 - $Array(\Delta X_2) = \{ \rightarrow,0,\rightarrow,\rightarrow,0,\rightarrow,0,0\}$.
So basically find the number of pairs of such binary strings where the initial substring of Body 1 $Array(\Delta X_1)$ always has lesser $\rightarrow$ than its counterpart's corresponding inital substring.
Is there a better approach? Hints please.