James the fly starts at (0, 0) Each step, James moves one unit right or one unit up. He is trying to get to the point (5, 7). However, at (4, 3) there is a frog that will eat him if he goes through that point. In how many ways can James reach (5, 7)?
+605 For obvious reasons, $(0,0)\to (m,n)$ can be done in $\binom{m+n}{m}=\binom{m+n}{n}$ ways. We let $f(m,n):=(0,0)\to (m,n)$.
So we just want $f(5,7)-f(4,3)f(5-4, 7-3)$, which is trivial to compute using the expression for $f(m,n)$.
+118677 James the fly starts at (0, 0) Each step, James moves one unit right or one unit up. He is trying to get to the point (5, 7). However, at (4, 3) there is a frog that will eat him if he goes through that point. In how many ways can James reach (5, 7)?
First I am going to consider how many ways assuming there is no fly eating frog
Each step must be up or right. There are 5 steps to the left and 7 steps up.
so there are 12C5 or 12C7 ways = 792ways
Now how many of these will be unfinished because the frog will get its supper?
that would be 7C3* 5C1 = 35*5 = 105 paths for the frog to get its supper.
So there are 792 - 105 = 687 safe routes.
