My question is trying to solve how many paths there are from $(x,y)$ to $(tx, ty)$ 3 possible moves are allowed at each step:
increment $x$ by $1$
increment $x$ by $2$
increment $y$ by $1$
I know that if only increments of 1 were allowed, then the solution is just:
$$\frac{((tx - x) + (ty - y))!}{((tx - x)!*(ty - y)!)}$$
But the incrementing x by two is really throwing me. Any hints would be appreciated. I'm not looking for the solution, just something to help me not be stumped.