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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)?

 May 15, 2021
 #1
avatar+556 
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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 $f(m,n)$.

 May 15, 2021
 #2
avatar+113625 
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I doubt that is 'obvious', or even makes any sense to over 98% of the people who use this forum.

Melody  May 15, 2021
edited by Melody  May 15, 2021
 #3
avatar+113625 
+1

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.

 May 15, 2021

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