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Difference between revisions of "2014 AMC 8 Problems/Problem 11"

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==Problem==
 
==Problem==
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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 avocado 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?
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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?
  
 
<math>\textbf{(A) }4\qquad\textbf{(B) }5\qquad\textbf{(C) }6\qquad\textbf{(D) }8\qquad \textbf{(E) }10</math>
 
<math>\textbf{(A) }4\qquad\textbf{(B) }5\qquad\textbf{(C) }6\qquad\textbf{(D) }8\qquad \textbf{(E) }10</math>

Revision as of 09:14, 2 December 2015

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?

$\textbf{(A) }4\qquad\textbf{(B) }5\qquad\textbf{(C) }6\qquad\textbf{(D) }8\qquad \textbf{(E) }10$

Solution

We can apply complementary counting and count the paths that DO go through the blocked intersection, which is $\dbinom{2}{1}\dbinom{3}{1}=6$. There are a total of $\dbinom{5}{2}=10$ paths, so there are $10-6=4$ paths possible. $\boxed{A}$ is the correct answer.

Solution 2

We can make a diagram of the roads available to Jack. [asy] size(50); defaultpen(linewidth(0.8)); pair A=(0,2),B=origin,C=(2,0),D=(2,2),E=(2,1),F=(3,2),G=(3,1),H=(3,0); draw(A--B--H--F--A); draw(D--C); draw(E--G); [/asy] Then, we can simply list the possible routes. [asy] size(50); defaultpen(linewidth(0.8)); pair A=(0,2),B=origin,C=(2,0),D=(2,2),E=(2,1),F=(3,2),G=(3,1),H=(3,0); draw(B--H--F); draw(D--C,dotted); draw(E--G,dotted); draw(F--A--B,dotted); [/asy] [asy] size(50); defaultpen(linewidth(0.8)); pair A=(0,2),B=origin,C=(2,0),D=(2,2),E=(2,1),F=(3,2),G=(3,1),H=(3,0); draw(B--C--E--G--F); draw(B--A--F,dotted); draw(D--E,dotted); draw(C--H--G,dotted); [/asy] [asy] size(50); defaultpen(linewidth(0.8)); pair A=(0,2),B=origin,C=(2,0),D=(2,2),E=(2,1),F=(3,2),G=(3,1),H=(3,0); draw(B--C--D--F); draw(B--A--D,dotted); draw(E--G,dotted); draw(F--H--C,dotted); [/asy] [asy] size(50); defaultpen(linewidth(0.8)); pair A=(0,2),B=origin,C=(2,0),D=(2,2),E=(2,1),F=(3,2),G=(3,1),H=(3,0); draw(B--A--F); draw(B--H--F,dotted); draw(D--C,dotted); draw(E--G,dotted); [/asy] There are 4 possible routes, so our answer is $\boxed{A}$.

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

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