Difference between revisions of "2014 AMC 8 Problems/Problem 11"
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We can apply complementary counting and count the paths that DO go through the blocked intersection, which is <math>\dbinom{2}{1}\dbinom{3}{1}=6</math>. There are a total of <math>\dbinom{5}{2}=10</math> paths, so there are <math>10-6=4</math> paths possible. <math>\boxed{A}</math> is the correct answer. | We can apply complementary counting and count the paths that DO go through the blocked intersection, which is <math>\dbinom{2}{1}\dbinom{3}{1}=6</math>. There are a total of <math>\dbinom{5}{2}=10</math> paths, so there are <math>10-6=4</math> paths possible. <math>\boxed{A}</math> is the correct answer. | ||
Revision as of 09:53, 3 November 2023
Contents
[hide]Problem
Jack wants to bike from his house to Jill's house, which is located three blocks east and two blocks north of Jack's house. After biking each block, Jack can continue either east or north, but he needs to avoid a dangerous intersection one block east and one block north of his house. In how many ways can he reach Jill's house by biking a total of five blocks?
Video Solution (CREATIVE THINKING)
~Education, the Study of Everything
Video Solution
https://www.youtube.com/watch?v=rxrQLNxESW0 ~David
https://youtu.be/pWLm41JtiCw ~savannahsolver
Solution 1
We can apply complementary counting and count the paths that DO go through the blocked intersection, which is . There are a total of paths, so there are paths possible. is the correct answer.
Solution 2
We can make a diagram of the roads available to Jack. Then, we can simply list the possible routes. There are 4 possible routes, so our answer is .
Note: This is not recommended in more complicated problems (e.g. Jill's house is 1000 blocks east and 400 blocks north of Jack's house).
See Also
2014 AMC 8 (Problems • Answer Key • Resources) | ||
Preceded by Problem 10 |
Followed by Problem 12 | |
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 AJHSME/AMC 8 Problems and Solutions |
The problems on this page are copyrighted by the Mathematical Association of America's American Mathematics Competitions.