Difference between revisions of "2013 AMC 8 Problems/Problem 21"

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(Solution)
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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 3C1=3 and the number of ways to get to school from there is 4C2=6 so the answer is <math>3\cdot 6 = \boxed{\textbf{(E)}\ 18}</math>.
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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 13:50, 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?

$\textbf{(A)}\ 3 \qquad \textbf{(B)}\ 6 \qquad \textbf{(C)}\ 9 \qquad \textbf{(D)}\ 12 \qquad \textbf{(E)}\ 18$

Solution

This is a good time to use constructive counting.

The number of ways to get to City Park is $\binom31 = 3$ and the number of ways to get to school from there is $\binom42=6$ so the answer is $3\cdot 6 = \boxed{\textbf{(E)}\ 18}$.

See Also

2013 AMC 8 (ProblemsAnswer KeyResources)
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

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