How many up-or-to-the-right paths from $(0,0)$ to $(8,5)$, and related questions Recall that in the section on functions, we created an encoding for the paths
from the point $(0, 0)$ to the point $(m, n)$ which either went right or up at each
step. In particular, we said that each path can be represented as a bit string of
length $m+ n$ with exactly $m$ 0's and $n$ 1's.
For this problem, let $m = 8, n = 5$.
a. How many paths are there from $(0, 0)$ to $(8, 5)$?
b. How many of these paths go through the point $(3, 3)$?
c. How many of them avoid the point $(3, 3)$?
 A: Hints:
a. How many ways are there of arranging eight $0$s and five $1$s in a string? 
b. How many ways are there of arranging three $0$s and three $1$s in a string? How many ways are there of arranging five $0$s and two $1$s in a string? How many ways are there of arranging three $0$s and three $1$s followed by five $0$s and two $1$s?
c. How is this related to a and b?  
A: I recommend googling the Binomial theorem, as it's an introduction to binomial coefficients. Going from (0,0) to (m,n) is one of the basics of probability. The answer to your question is; $\frac {n!} {k!*(n-k)!}$; n choose k

n in this case would be te sum of the lines needed to make a path; m+n,
and k would be either horizontal of vertical, so m or n
$\frac {n!} {k!*(n-k)!}$ becomes $\frac {(m+n)!} {m!*n!}$
For things like "going through (3,3)", you could use m = 3, n = 3 to go to (3,3) and multiply that by "going from (3,3) to (8-3, 5-3)", which is the same as using $\frac {(m+n)!} {m!*n!}$ again with m = 8-3, n = 5-3
