Difference between revisions of "2010 AMC 12A Problems/Problem 18"

(Solution)
(Solution)
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Hence the total number of paths is <math>2(1+64+784) = \boxed{1698}</math>.
 
Hence the total number of paths is <math>2(1+64+784) = \boxed{1698}</math>.
<math>{1024\choose512\choose256\choose128\choose64\choose32\choose16\choose8\choose4\choose2\choose1}</math>
+
<math>{1024\choose{512\choose{256\choose{128\choose{64\choose{32\choose{16\choose{8\choose{4\choose{2\choose{1}}}}}}}}}}}</math>
  
 
== See also ==
 
== See also ==

Revision as of 20:48, 5 June 2015

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 are $1$ paths 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 $2(1+64+784) = \boxed{1698}$. ${1024\choose{512\choose{256\choose{128\choose{64\choose{32\choose{16\choose{8\choose{4\choose{2\choose{1}}}}}}}}}}}$

See also

2010 AMC 12A (ProblemsAnswer KeyResources)
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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