Q:

The number of cases $(0, 0)$ moves by either $(1,1)$ or $(1,-1)$, in $2n$ steps, without touching $x$-axis again.

I was solving combinatorics problems when I ran into this shady statement:
The number of cases $(0, 0)$ moves by either $(1, 1)$ or $(1, -1)$, in $2n$ steps, without touching $x$-axis again is $\binom{2n}{n}$.
I tried in vain to understand this result. It produces a modified Pascal's triangle and it gives the right answer. How to prove this statement using a combinatorical proof?

A:

Let's focus on the scenario where $Y_1=1$. The key is to use refection principle on the next $2n-1$ moves.
There will be 4 all inclusive and mutually exclusive scenarios:

$Y_i>0$ for all $i$
$Y_n=0$, of course x-axis must have been touched
$Y_n<0$, of course x-axis must have been touched
$Y_n>0$ and $\exists i$ s.t. $Y_i=0$, i.e. x-axis is touched

Let $N_i$ be the number of paths that fall in scenario i. 
$N_2=C_{2n-1}^n$
$N_3=\sum_{k=n+1}^{2n-1}C_{2n-1}^k$
With reflection principle, we have $N_3=N_4$.
Since $N_1+N_2+N_3+N_4=2^{2n-1}$, 
$N_1=2^{2n-1}-C_{2n-1}^n-2\sum_{k=n+1}^{2n-1}C_{2n-1}^k=\sum_{k=1}^{2n-1}C_{2n-1}^k-C_{2n-1}^n-\sum_{k=n+1}^{2n-1}C_{2n-1}^k-\sum_{k=1}^{n-2}C_{2n-1}^k=C_{2n-1}^{n-1}=\frac{(2n-1)!}{(n-1)!n!}=\frac{(2n)!}{2(n!)^2}=\frac{C_{2n}^n}{2}$
Now adding the part where $Y_1=-1$, the total is $C_{2n}^n$.

