[Extension of this]
We can move in 4-directions and we need to reach $(0,b)$ from $(a,0)$ in exactly $n$ steps keeping in the first quadrant ($x\ge0$ and $y\ge0$) [$a,b\ge0$]
Similar to previous problem, $a+n_r\ge n_l$, $n_u\ge n_d$ and finally $a+n_r-n_l=0$, $n_u-n_d=b$, $n_u+n_d+n_l+n_r=n$. (let $n_u=k$, a similar logic as this is used below)
$$\sum_{k=0}^{\frac{n+a+b}2}\binom n{2k-b}\left(\binom{2k-b}{k}-\binom{2k-b}{k+1}\right)\left(\binom{n+b-2k}{\frac{n+a+b}2-k}-\binom{n+b-2k}{\frac{n+a+b}2-k+1}\right)\\
=\sum_{k=0}^{\frac{n+a+b}2}\binom n{2k-b}\frac{(b+1)(a+1)}{(2k-b+1)(n+b-2k+1)}\binom{2k-b+1}{k+1}\binom{n+b-2k+1}{\frac{n+a+b}2-k+1}\\
=\frac{(a+1)(b+1)}{(n+1)(n+2)}\binom{n+2}{\frac{n-a+b}2+1}\sum_{k=0}^{\frac{n+a+b}2}\binom{\frac{n-a+b}2+1}{k+1}\binom{\frac{n+a+b}2+1}{k-b}\\
=\frac{(a+1)(b+1)}{(n+1)(n+2)}\binom{n+2}{\frac{n-a+b}2+1}\binom{n+2}{\frac{n-a-b}2}$$
And at $a=0,b=0$:
$$\frac{1}{(n+1)(n+2)}\binom{n+2}{\frac{n}2+1}\binom{n+2}{\frac{n}2}$$
the solution seems to agree with previous result.
Question: It seems correct. What do you say? Is there any easy way?
