How many paths are there from $(0,0)$ to $(20,10)$ adding $(0,1)$ (up) or $(1,0)$ (right) component-wise? How many paths are there from $(0,0)$ to $(20,10)$ adding $(0,1)$ (up) or $(1,0)$ (right) component-wise?
I've broken this down into $10$ ups and $20$ rights are needed in $30$ positions.
Ex. (up, right, up, ..., up) where $\sum up = 10 \land \sum right = 20$.
Since this seems to be the equivalent to the number of ways you can order $10$ similar objects into $30$ positions or $20$ similar objects in $30$ positions, the number of paths should be $_{30}C_{10}$.  Does this make sense or am I double counting things?
 A: Looks like you have already figured it out, with some caveats. 
Since you have discretized the grid you can treat it like a counting problem. If we assume you only care about the most efficient paths (those with exactly 10 ups and 20 rights), then you care about the number of different combinations of "right" or "up" you can put together. This means your path has to be 30 steps long, and anything that isn't "up" is "right". Since you can't tell the "ups" apart from each other, it doesn't matter what order they are in. (After all you can't tell "up, up, right, right" from "up, up, right, right")
Also, since you are only concerned with ups and rights, you can't use more than 10 ups and 20 rights (because then you couldn't connect to (20,10)). Therefore you are also without replacement. So you are fixed to 30 selections, have only 2 options for each of the 30 spaces, the order doesn't matter, and there's no replacement. That is a textbook example of combination (https://en.wikipedia.org/wiki/Combination#Number_of_combinations_with_repetition). The combination would be counted as $30 \choose {10}$ = $30 \choose 20$, as you said above.
The caveat which I think you have also noticed from the way the question is phrased is that there are many more paths if you also allow for "downs" and "lefts". If you allow these there are an infinite number of paths. Even if you constrain the path to lie inside the rectangle between (0,0) and (20,10), because the path could loop on itself using "right,down, left, up" an arbitrary number of times.
An interesting follow up question might be: How many paths are there from (0,0) to (20,10) using up,down,right, left that cannot contact the same grid point twice? 
