Number of journeys from origin to origin using 2n moves.

Question

Suppose I must use $2n$ moves to get from (0, 0) to (0, 0) using vectors $\pmatrix {0 \\ 1}$, $\pmatrix {1 \\ 0}, \pmatrix{0 \\ -1}, \pmatrix{-1 \\ 0}$.
And I want to find out how many ways are there to choose such a journey. So we must have $w\pmatrix {1 \\0} + x\pmatrix{0 \\ 1} +y\pmatrix{0 \\ - 1} +z\pmatrix{-1 \\ 0}= \pmatrix{0 \\ 0}$, for some positive integers w,x,y,z.
Where we have $w+x+y+z = 2n$, and more specifically $w -z = 0$, and $x-y = 0 $. And we have $w=z, x=y \implies z+x=n$, since $0 \leq z \leq n$, we have $n+1$ pairs for (z,x), and more specifically $n+1$ options for (w,x,y,z). But now I am stuck at how to find how many journeys there are, and do not know how to approach this. One idea was to use partitions of $z$ and $x$, but the number of partitions of a certain integer is found by a recurrence relation, and if n is very large, it's not viable to calculate the number of journeys.

Answer 1

Condition on the number $k$ of steps in the $(1,0)$ direction, and apply Vandermonde: $$ \sum_{k=0}^n \binom{2n}{k,k,n-k,n-k} = \sum_{k=0}^n \binom{2n}{n}\binom{n}{k}^2 = \binom{2n}{n} \sum_{k=0}^n \binom{n}{k}^2 = \binom{2n}{n}^2 $$