Talk:2021 Fall AMC 12B Problems/Problem 17

Solution 3

Use generating function, define $c_{n}\cdot x^{n}$ 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: $p(x)=6x$ \r\n \\2nd step: $p(x)=6x\cdot(x^{-1}+2) = 6 + 12x$ \\3rd step: $p(x)=6\cdot6x + 12x\cdot(x^{-1}+2) = 12 + 60x$ \\4th step: $p(x)=12\cdot6x + 60x\cdot(x^{-1}+2) = 60 + 192x$ \\5th step: $p(x)=60\cdot6x + 192x\cdot(x^{-1}+2) = 192 + 744x$

\\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

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Talk:2021 Fall AMC 12B Problems/Problem 17

Solution 3