Show that the number of walks from $(0, 0)$ to $(2n, 0)$ that are positive is equal to $\frac{1}{2n-1}{2n-1 \choose n}$.
(Such a walk is called a bridge.)
Here, one step within the walk is a movement of one unit to the right, with the possibility of moving up for $1$ and $-1$. I believe that the walk being positive means it stays above the x-axis and only touches it at $(0,0)$ and $(2n,0)$
I have been thinking that this is the same number of paths from $(1,1)$ to $(2n-1, 1)$ that do not touch the x-axis but I am not sure if this is the right way to go about it.
What you need is the Catalan Numbers, which count the "non-negative" walks from $(0,0)$ to $(2n,0)$. They are given by $C_n = \frac{1}{n+1} \binom{2n}{n}$
Now as we want only positive walks the first move has to be to $(1,1)$, while the last one has to be from $(2n-1,1)$ to $(2n,0)$. So the first and the last move are fixed and we need to find the number of non-negative walks from $(1,1)$ to $(2n-1,1)$, which is same as the non-negative walks from $(0,0)$ to $(2(n-1),0)$. That number is given by:
$$C_{n-1} = \frac 1n \binom{2n-2}{n-1} = \frac{(2n-2)!}{(n-1)! \cdot n!} = \frac1{2n-1}\frac{(2n-1)!}{(n-1)!n!} = \frac{1}{2n-1}\binom{2n-1}{n}$$