Discrete Mathematics Riddle, square grid problem I've been trying to solve this problem in my book to no avail: 



I am even stuck on part a... to find a closed form of this I started with size 2x2 square, counted the number of paths, then used a 3x3 square, did the same but I feel that was a rather inefficient way to go about doing this problem. Can someone walk me through the solutions for parts (a), (b), (c)? Thanks.
You can write the number of ways to get to each intersection point while minimizing Taxicab distance by using the fact that the number of paths at any given intersection is equal to the sum of the number of paths that come directly from the left and the number of paths that come directly from below.



You might notice that this looks like Pascal's Triangle turned on its side. That's because it is exactly that. The reason these numbers are equal to binomial coefficients is because the number of paths is exactly the number of ways to arrange the rights and lefts in the path; to get to point $B$ you need $4$ lefts and $4$ rights. There are $$\binom{8}{4}=70$$ ways to choose the order of these rights and lefts. 



You should be able to solve the second question using this method.

Good luck :)



EDIT: I've included how the numbers would be assigned for the third part. Notice the $10$ assigned to $(2,4)$.
For part (a), you have in total 8 moves to make, 4 of which has to be up and the other 4 right. You can rearrange the order of these moves in $\binom{8}{4} = 70$ ways. So that is the answer to part (a).

For part (b), you have in total $n+m$ moves to make, n of which has to be up and the other m right. You can rearrange the order of these moves in $\binom{n+m}{n}$ ways.