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

m (→‎Solution: Added direction of walking)
(→‎Solution: straaange wording. added diagram.)
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==Solution==
 
==Solution==
−
This is a good time to use constructive counting.
+
<asy>
 +
unitsize(8mm);
 +
for(int i=0; i<=8; ++i)
 +
{
 +
draw((0,i)--(8,i));
 +
draw((i,0)--(i,8));
 +
}
 +
fill((2,1)--(6,1)--(6,6)--(2,6)--cycle);
 +
for(int j=0; j<= 38; ++j)
 +
{
 +
draw((0,0)--(2,0)--(2,1)--(0,1)--(0,0), black+linewidth(3));
 +
draw((6,6)--(6,8)--(8,8)--(8,6)--(6,6), black+linewidth(3));
 +
}</asy>
  
−
The number of ways to get from Samantha to City Park is <math>\binom31 = 3</math>, and the number of ways to get to school from City Park is <math>\binom42=6</math> so the answer is <math>3\cdot 6 = \boxed{\textbf{(E)}\ 18}</math>.
+
The number of ways to get from Samantha's house to City Park is <math>\binom31 = 3</math>, and the number of ways to get from City Park to school is <math>\binom42=6</math>. Since there's one way to go through City Park (just walking straight through), the number of different ways to go from Samantha's house to City Park to school <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, 29 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

[asy] unitsize(8mm); for(int i=0; i<=8; ++i) { draw((0,i)--(8,i)); draw((i,0)--(i,8)); } fill((2,1)--(6,1)--(6,6)--(2,6)--cycle); for(int j=0; j<= 38; ++j) { draw((0,0)--(2,0)--(2,1)--(0,1)--(0,0), black+linewidth(3)); draw((6,6)--(6,8)--(8,8)--(8,6)--(6,6), black+linewidth(3)); }[/asy]

The number of ways to get from Samantha's house to City Park is $\binom31 = 3$, and the number of ways to get from City Park to school is $\binom42=6$. Since there's one way to go through City Park (just walking straight through), the number of different ways to go from Samantha's house to City Park to school $3\cdot 6 = \boxed{\textbf{(E)}\ 18}$.

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

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