Permutations for paths 
What is the counting sequence for paths from $$(0,0)\text{ to }(n,n)$$ where $$2n$$ is the size of the path (number of steps) and n can vary over all nonnegative integers.

I don't know how to approach this problem. I know that in general, # of paths from $(a,b)$ to $(c,d)$ is $(m+n)!\,/(m!n!)$ where m and n denote $(c-a)$ and $(d-b)$ respectively. But this size of the path $2n$ confuses me.
 A: In general, the number of paths from $(a,b)$ to $(c,d)$ given $c \geq a$ and $d \geq b$ is
$$\binom{m+n}{n} = \binom{m+n}{m} = \frac{(m+n)!}{m!\cdot n!} $$
where the steps go up and to the right only. (Otherwise our paths might be arbitrarily long.)
So for your question, a path going from $(0,0)$ to $(n,n)$ where the steps go up and to right is equivalent to a sequence of $2n$ steps consisting of $n$ steps to the right and $n$ steps up. 
Once we have chosen the position of the steps up, the steps to the right are all forced so the number of paths from $(0,0)$ to $(n,n)$ is $\binom{2n}{n}$.
A: The information about the size implies that you go only $\rightarrow, \uparrow.$
Hence, the answer is 
$$\binom{2n}{n}=\frac{(2n)!}{n!n!}.$$
This represents that you choose $n$ $(\rightarrow s)$ in $2n$ steps. Of course, the order matters.
For example, when $n=3$, what you want is the number of permutations of $\rightarrow \rightarrow \rightarrow \uparrow\uparrow\uparrow$, which is $6!/(3!\cdot 3!).$
A: I guess information about size path is redundant and can be simply ignored.
Length (or size) of the shortest path from (0,0) to (n,n) IS 2n.
Your approach is correct and the answer would be (2n)!/(n!n!)
