Talk:2021 Fall AMC 12B Problems/Problem 17

Revision as of 14:12, 28 November 2021 by Wwei.yu (talk | contribs) (Solution 3)

Solution 3

Use generating function, define ${c(n)}*x^{n}$ be ${c(n)}$ ways for the end point be ${n}$ unit away from the origins. Therefore, if the current point is origin, need to $\cdot6{x}$ if the current point on vertex of the unit hexagon, need to $\cdot(x^{-1}+2)$, where there is one way to return to the origin and there are two ways to keep distance = 1

Now let's start init $p(x)=1$ 1st step: $p(x)=6x$ 2nd step: $p(x)=6x*(x^{-1}+2) = 6 + 12x$ 3rd step: $p(x)=6*6x + 12x*(x^{-1}+2) = 12 + 60x$ 4th step: $p(x)=12*6x + 60x*(x^{-1}+2) = 60 + 192x$ 5th step: $p(x)=60*6x + 192x*(x^{-1}+2) = 192 + 744x$

So there are $192+744=936$ ways for the bug never moves more than 1 unit away from orign, and $\frac{936}{6^5} = \boxed{\frac{13}{108}}.$

-wwei.yu