North-east lattice path ($E=(1,0),N=(0,1)$)

A north-east lattice path is a lattice path whose allowed steps are $E=(1,0)$ and $N=(0,1)$. From $(0,0)$ to $(a,b)$, with nonnegative integers $a,b$, it has $a$ east steps and $b$ north steps. The number isThere is one zero-step north-east lattice path when $a=b=0$.^[ref]

Seven positions, thirty-five paths. To reach $(4,3)$, choose the three positions occupied by north steps in a seven-symbol word. For example, the subset $\{2,4,7\}$ of $\{1,2,3,4,5,6,7\}$ gives ENENEEN: putting $N$ exactly in those positions and $E$ elsewhere reconstructs the north-east lattice path. Reading the north-step positions recovers the subset, so these two operations give a bijection. HenceThe numerator counts ordered choices of three different positions; each subset occurs in $3!=6$ orders. Dividing by this factorial counts each north-east lattice path once. A binary string recording $N$ as $1$ and $E$ as $0$ gives the same example as 0101001.

This interpretation is useful when arrangements have fixed totals but flexible order: a subset, a binary string, and a north-east lattice path become different views of one selection. It also separates endpoint restrictions from restrictions along the journey. Requiring the lattice path to avoid specified points changes the count; the unrestricted binomial coefficient no longer suffices, but the local north-east lattice path recurrence still suggests how to count allowed arrivals.

For positive integers $a,b$, the north-east lattice path recurrence isThe boundaries have only straight north-east lattice paths. The origin counts the zero-step north-east lattice path once.^[ref]

Proof outline. A north-east lattice path ending at $(a,b)$ has a final step either from $(a-1,b)$ or from $(a,b-1)$. Deleting that step gives a shorter north-east lattice path; appending it reverses the operation. The two classes are disjoint and exhaust all possibilities, so the rule of sum adds their counts. At $(4,3)$, the arrivals from $(3,3)$ contribute $\left(\genfrac{}{}{0ex}{}{6}{3}\right)=20$, while those from $(4,2)$ contribute $\left(\genfrac{}{}{0ex}{}{6}{2}\right)=15$: altogether $20+15=35$.

Set $P(a,b)=0$ if either integer coordinate is negative. Then the same recurrence relation applies on the nonnegative axes away from $(0,0)$: one predecessor contributes one, and the other contributes zero. Initialize $P(0,0)=1$ separately; applying the recurrence relation there would incorrectly give zero.

Computing the grid once in increasing order of $a+b$ is dynamic programming: every entry uses two earlier entries. If a specified point is forbidden, set its count to zero, then continue the same recurrence relation at allowed points. This counts north-east lattice paths avoiding the forbidden points, provided the origin and endpoint are allowed. The last-step argument remains valid because appending a step to an allowed endpoint cannot visit any new intermediate integer lattice point.