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Difference between revisions of "Talk:2021 Fall AMC 12B Problems/Problem 17"

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Latest revision as of 17:52, 28 November 2021

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 with $p(x)=1$;

1st step: $p(x)=6x$

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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