Suppose we start at the origin and we're interested in how many paths there are such that we only return back to the origin after m steps and not before. where m is some even integer.
Each step can be up one unit or down one unit. Suppose for simplicity we can only have positive values. Then if m=8, one path could be (1,2,3,2,3,2,1,0) another could be (1,2,3,4,3,2,1,0).
I dont see how to remove paths that return to the origin too soon from my count of possible paths.
On
You can use slightly modified Catalan Numbers. Note that the first move has to be $0 \to 1$, while the last one has to be $1 \to 0$. Then we need to determine the other $m-2$ moves, s.t. we never go below $1$ and starting at $1$ we return to $1$. This number is exactly the $(m-2)$-th Catalan Number. So the wanted number is:
$$C_{m-2} = \frac1{m-1}\binom{2m-4}{m-2}$$
Hint: Consider $(u+d)^{2k}$ and try to find the coefficient of $u^kd^k$, which will correspond to the number of ways to obtain a path with $k$ total steps up and $k$ total steps down, so that you have indeed returned to the origin.