Difference between revisions of "2010 AMC 12A Problems/Problem 18"
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== Solution 2== | == 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 <math>S</math>). There is symmetry about <math>y=x</math>, so we only have to consider <math>(1,-1), (1,0)</math>, and <math>(1,1)</math>. <math>(1,1)</math> can go on the boundary in two ways, so we can only consider one case and multiply it by two. For <math>(1,0)</math> and <math>(1,-1)</math> we can just multiply by two. So we count paths from <math>(-4,4)</math> 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 <math>(4,4)</math>, and all in all, we get the answer is <math>\dbinom{16}{8}-2[\dbinom{8}{3}\dbinom{7}{2}+\dbinom{9}{4}\dbinom{6}{2}+\dbinom{10}{5}\dbinom{5}{2}]=1698</math>, which is answer choice <math>\textbf{(D)}</math> | 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 <math>S</math>). There is symmetry about <math>y=x</math>, so we only have to consider <math>(1,-1), (1,0)</math>, and <math>(1,1)</math>. <math>(1,1)</math> can go on the boundary in two ways, so we can only consider one case and multiply it by two. For <math>(1,0)</math> and <math>(1,-1)</math> we can just multiply by two. So we count paths from <math>(-4,4)</math> 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 <math>(4,4)</math>, and all in all, we get the answer is <math>\dbinom{16}{8}-2[\dbinom{8}{3}\dbinom{7}{2}+\dbinom{9}{4}\dbinom{6}{2}+\dbinom{10}{5}\dbinom{5}{2}]=1698</math>, which is answer choice <math>\textbf{(D)}</math> | ||
− | -vsamc | + | -vsamc |
+ | Note: Sorry if this was rushed. | ||
== See also == | == See also == | ||
{{AMC12 box|year=2010|num-b=17|num-a=19|ab=A}} | {{AMC12 box|year=2010|num-b=17|num-a=19|ab=A}} |
Revision as of 11:04, 11 March 2020
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
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 , , and .
There is path of the first kind, paths of the second kind, and paths of the third type. Each path that goes through the fourth quadrant must pass through exactly one of the points , , and . Again, there is path of the first kind, paths of the second kind, and paths of the third type.
Hence the total number of paths is .
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 ). There is symmetry about , so we only have to consider , and . can go on the boundary in two ways, so we can only consider one case and multiply it by two. For and we can just multiply by two. So we count paths from 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 , and all in all, we get the answer is , which is answer choice -vsamc Note: Sorry if 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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