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2010 AMC 12A Problems/Problem 18

Revision as of 10:03, 11 March 2020 by Vsamc (talk | contribs)

Problem

A 16-step path is to go from $(-4,-4)$ to $(4,4)$ with each step increasing either the $x$-coordinate or the $y$-coordinate by 1. How many such paths stay outside or on the boundary of the square $-2 \le x \le 2$, $-2 \le y \le 2$ at each step?

$\textbf{(A)}\ 92 \qquad \textbf{(B)}\ 144 \qquad \textbf{(C)}\ 1568 \qquad \textbf{(D)}\ 1698 \qquad \textbf{(E)}\ 12,800$

Solution

Each path must go through either the second or the fourth quadrant. Each path that goes through the second quadrant must pass through exactly one of the points $(-4,4)$, $(-3,3)$, and $(-2,2)$.

There is $1$ path of the first kind, ${8\choose 1}^2=64$ paths of the second kind, and ${8\choose 2}^2=28^2=784$ paths of the third type. Each path that goes through the fourth quadrant must pass through exactly one of the points $(4,-4)$, $(3,-3)$, and $(2,-2)$. Again, there is $1$ path of the first kind, ${8\choose 1}^2=64$ paths of the second kind, and ${8\choose 2}^2=28^2=784$ paths of the third type.

Hence the total number of paths is $2(1+64+784) = \boxed{1698}$.

Solution 2

We will use the concept of complimentary counting. If you go in the square you have to go out by these labeled points(click the link) https://imgur.com/VysX4P0 and go out through the borders because if you didnt, you would touch another point in one of those points in the set of points in the link(Call it $S$). There is symmetry about $y=x$, so we only have to consider $(1,-1), (1,0)$, and $(1,1)$. $(1,1)$ can go on the boundary in two ways, so we can only consider one case and multiply it by two. For $(1,0)$ and $(1,-1)$ we can just multiply by two. So we count paths from $(-4,4)$ to each of these points, and then multiply that by the number of ways to get from the point one unit right of that to $(4,4)$, and all in all, we get the answer is $\dbinom{16}{8}-2[\dbinom{8}{3}\dbinom{7}{2}+\dbinom{9}{4}\dbinom{6}{2}+\dbinom{10}{5}\dbinom{5}{2}]=1698$, which is answer choice $\textbf{(D)}$ -vsamc(sorry this was rushed)

See also

2010 AMC 12A (Problems • Answer Key • Resources)
Preceded by
Problem 17 		Followed by
Problem 19
1 • 2 • 3 • 4 • 5 • 6 • 7 • 8 • 9 • 10 • 11 • 12 • 13 • 14 • 15 • 16 • 17 • 18 • 19 • 20 • 21 • 22 • 23 • 24 • 25
All AMC 12 Problems and Solutions

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