2010 AMC 12A Problem 18

Every step increases $x+y$ by $1,$ which runs from $-8$ to $8,$ so each path passes through exactly one lattice point with $x+y=0.$ To stay out of the open square, that point $(t,-t)$ must have $|t|\ge2,$ so it is one of $(\pm2,\mp2),(\pm3,\mp3),(\pm4,\mp4).$ By symmetry consider the three points $(-4,4),(-3,3),(-2,2)$ and double. The number of paths from $(-4,-4)$ to $(-(4-j),4-j)$ is $\binom{8}{j},$ and the number continuing on to $(4,4)$ is also $\binom{8}{j}.$ Therefore the total is $$\begin{gathered} 2\left(\binom80^2+\binom81^2+\binom82^2\right) \\ =2(1+64+784) \\ =1698. \end{gathered}$$ Thus, D is the correct answer.