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Mack the bug starts at $(0,0)$ at noon and each minute moves one unit right or one unit up. He is trying to get to the point $(5,7)$. However, at $(2,3)$ there is a spider that will eat him if he goes through that point. In how many ways can Mack reach $(5,7)$?

Guest Mar 20, 2015
 #1
avatar+92674 
+5

The number of ways to get from (0, 0) to (5, 7) moving in this manner is given by

(7 + 5) ! / (7! 5!) = 12! / (7! 5! )  = 792 ways

From these, we want to subtract the ways to get from (0, 0) to (2, 3)

This is given by 

(3 + 2)! / (3! 2!) = 5! / (3! 2!) =  10 ways

So 792 - 10  = 782 routes

 

  

CPhill  Mar 20, 2015
 #2
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If you would explain one of the factorials it would be greatly appreciated. I don't really understand how they are generated ,say, for the number of routes to the spider.

Guest Mar 20, 2015
 #3
avatar+92674 
+5

I'm gonna' be honest, Anonymous....I "borrowed" this "formula" from one of our fellow posters, Nauseated, based on his answer to a problem just last week....I haven't had time to digest the rationale behind it, yet.

However.......here is his "non-technical" explanation.....if you care to look at it....

{It's the fourth answer from the bottom}.........http://web2.0calc.com/questions/each-small-square-has-a-side-length-of-one-unit-how-many-distinct-paths-of-six-units-are-there-from-a-to-b

 

  

CPhill  Mar 20, 2015
 #5
avatar+92674 
0

Thanks for that link, Anonymous.....that really helped me see the logic behind the formula!!!

 

  

CPhill  Mar 21, 2015

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