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

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<math>\textbf{(A)}\ 3 \qquad \textbf{(B)}\ 6 \qquad \textbf{(C)}\ 9 \qquad \textbf{(D)}\ 12 \qquad \textbf{(E)}\ 18</math>
 
<math>\textbf{(A)}\ 3 \qquad \textbf{(B)}\ 6 \qquad \textbf{(C)}\ 9 \qquad \textbf{(D)}\ 12 \qquad \textbf{(E)}\ 18</math>
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==Video Solution==
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https://youtu.be/OOdK-nOzaII?t=2679
  
 
==Solution==
 
==Solution==
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}</asy>
 
}</asy>
  
We see that Samantha only has one way to go through in the city park, 6 ways to the city park to school, and 3 ways from her house to the city park.  After this, we can see that there are <math>6 \cdot 3 = 18</math>ways for Samantha to go to school. Hence, <math>\textbf{E}</math>
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Using [[combinations]], we get that the number of ways to get from Samantha's house to City Park is <math>\binom31 = \dfrac{3!}{1!2!} = 3</math>, and the number of ways to get from City Park to school is <math>\binom42= \dfrac{4!}{2!2!} = \dfrac{4\cdot 3}{2} = 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 14:05, 6 October 2020

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?

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

Video Solution

https://youtu.be/OOdK-nOzaII?t=2679

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]

Using combinations, we get that the number of ways to get from Samantha's house to City Park is $\binom31 = \dfrac{3!}{1!2!} = 3$, and the number of ways to get from City Park to school is $\binom42= \dfrac{4!}{2!2!} = \dfrac{4\cdot 3}{2} = 6$. 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 (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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