Difference between revisions of "2010 AMC 12A Problems/Problem 18"
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| − | == Problem | + | == Problem == |
A 16-step path is to go from <math>(-4,-4)</math> to <math>(4,4)</math> with each step increasing either the <math>x</math>-coordinate or the <math>y</math>-coordinate by 1. How many such paths stay outside or on the boundary of the square <math>-2 \le x \le 2</math>, <math>-2 \le y \le 2</math> at each step? | A 16-step path is to go from <math>(-4,-4)</math> to <math>(4,4)</math> with each step increasing either the <math>x</math>-coordinate or the <math>y</math>-coordinate by 1. How many such paths stay outside or on the boundary of the square <math>-2 \le x \le 2</math>, <math>-2 \le y \le 2</math> at each step? | ||
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For example: | For example: | ||
| − | < | + | <cmath>\begin{align*}&(-4,-4) \Rightarrow 1\\ |
| − | + | &(-3,-4) \Rightarrow 1\\ | |
| − | + | &(-2,-4) \Rightarrow 1\\ | |
| − | + | &(-4,-3) \Rightarrow 1\\ | |
| − | + | &(-4,-2) \Rightarrow 1\\ | |
| − | + | &(-3,-3) \Rightarrow 2\\ | |
| − | + | &(-2,-3) \Rightarrow 3\\ | |
| − | + | &(-3,-2) \Rightarrow 3\\ | |
| − | + | &(-2,-2) \Rightarrow 6\end{align*}</cmath> | |
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etc. | etc. | ||
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There is exactly <math>1</math> path of the first kind, <math>{8\choose 1}^2=64</math> paths of the second kind, and <math>{8\choose 2}^2=28^2=784</math> paths of the third type. The conclusion remains the same. | There is exactly <math>1</math> path of the first kind, <math>{8\choose 1}^2=64</math> paths of the second kind, and <math>{8\choose 2}^2=28^2=784</math> paths of the third type. The conclusion remains the same. | ||
| + | |||
| + | == See also == | ||
| + | {{AMC12 box|year=2010|num-b=17|num-a=19|ab=A}} | ||
| + | |||
| + | [[Category:Introductory Combinatorics Problems]] | ||
Revision as of 23:32, 25 February 2010
Contents
Problem
A 16-step path is to go from 
to 
with each step increasing either the 
-coordinate or the 
-coordinate by 1. How many such paths stay outside or on the boundary of the square 
, 
at each step?
Solution
Brute Force Solution
The number of ways to reach any point 
on the grid is equal to the number of ways to reach 
plus the number of ways to reach 
. Using this recursion, we can draw the diagram and label each point with the number of ways to reach it and go up until we reach the end. Luckily, the figure is not so big that this is too time-consuming or difficult to do.
For example:
etc.
We soon reach 
Combinatorial Solution 1
By symmetry we only need to count the paths that go through the second quadrant (
, 
).
For each of these paths, let 
be the first point when it reaches 
. Clearly 
and the previous point on such path has to be 
.
Fix the value of 
. There are 
ways how the path can go from to , and then there are 
ways how the path can go from to .
Hence for 
we get 
paths, for 
we get 
paths, and for 
we get 
paths. This gives us 
paths through the second quadrant, hence the total number of paths is 
.
Combinatorial Solution 2
Each path that goes through the second quadrant must pass through exactly one of the points 
, 
, and 
.
There is exactly 
path of the first kind, 
paths of the second kind, and 
paths of the third type. The conclusion remains the same.
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 | |
