Lattice Paths problem I was assigned to determine the number of "lattice paths" that are in a 11 x 11 square. 
Recalling that I can only go upwards and rightwards, here is my approach:
Note: The red square is the restricted area I cannot go through.

I computed the result in WolframAlpha, and I got the following:
(Corrected input)

I just know that we go up to 11 choose 3 since we have 11 paths (n), and then we have up to 3 m moves. I really do not know what is the technical approach to this. I only know that by intuition, is up to 11 choose 3. Can someone explain to me why? 
Also, I want to know if my approach to the problem is correct.
Thank you!

Your solution is okay.
$$2\times\left[\binom{11}{0}\binom{11}{11}+\binom{11}{1}\binom{11}{10}+\binom{11}{2}\binom{11}{9}+\binom{11}{3}\binom{11}{8}\right]$$
If you go 'right-under' then it is for
certain that you will arrive $\left(11,0\right)$, $\left(10,1\right)$,
$\left(9,2\right)$ or $\left(8,3\right)$. 
It cannot happen that you arrive on your path on more than one of
these points.
The factor $2$ covers the fact that you can also go 'left-up' along analogous paths.

Edit:
Your approach is correct.
Going from $(0,0)$ to e.g. $(9,2)$ with steps to the right or upwards comes to choosing $2$ (numbered) steps out of $9+2=11$ to be the steps that go upward. There are $\binom{11}{2}$ ways to do that. Of course it also comes to choosing $9$ out of $11$ to be the steps to the right, giving $\binom{11}{9}$ possibilities and (fortunately :)) $\binom{11}{2}=\binom{11}{9}$.
