Difference between revisions of "2013 AMC 8 Problems/Problem 21"
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This is a good time to use constructive counting. | This is a good time to use constructive counting. | ||
| − | The number of ways to get to City Park is | + | The number of ways to get to City Park is <math>\binom31 = 3</math> and the number of ways to get to school from there is <math>\binom42=6</math> so the answer is <math>3\cdot 6 = \boxed{\textbf{(E)}\ 18}</math>. |
==See Also== | ==See Also== | ||
{{AMC8 box|year=2013|num-b=20|num-a=22}} | {{AMC8 box|year=2013|num-b=20|num-a=22}} | ||
{{MAA Notice}} | {{MAA Notice}} | ||
Revision as of 12:50, 27 November 2013
Problem
Samantha lives 2 blocks west and 1 block south of the southwest corner of City Park. Her school is 2 blocks east and 2 blocks north of the northeast corner of City Park. On school days she bikes on streets to the southwest corner of City Park, then takes a diagonal path through the park to the northeast corner, and then bikes on streets to school. If her route is as short as possible, how many different routes can she take?
Solution
This is a good time to use constructive counting.
The number of ways to get to City Park is 
and the number of ways to get to school from there is 
so the answer is 
.
See Also
| 2013 AMC 8 (Problems • Answer Key • Resources) | ||
| Preceded by Problem 20 |
Followed by Problem 22 | |
| 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. 
