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Difference between revisions of "2014 AMC 8 Problems/Problem 11"
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<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>
 +
 +
==Insight==
 +
This is a combonatorics problem. For these problems, you should start by counting the total, then subtracting the ways that don't work. Using word rearrangements, we can find the total number of paths, then find the number of paths through the dangerous intersection using the same method. Then you subtract the first from the second. This is called Complementary Counting
 +
-JasonDaGoat
 +
 +
==Video Solution (CREATIVE THINKING)==
 +
https://youtu.be/YCr5GFcmdcI
 +
 +
~Education, the Study of Everything
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 +
 +
 +
 +
==Video Solution==
 +
https://www.youtube.com/watch?v=rxrQLNxESW0  ~David
 +
 +
https://youtu.be/pWLm41JtiCw ~savannahsolver
 +
 +
==Solution 1==
 +
We can apply complementary counting and count the paths that DO go through the blocked intersection, which is <math>\dbinom{2}{1}\dbinom{3}{1}=6</math>. There are a total of <math>\dbinom{5}{2}=10</math> paths, so there are <math>10-6=4</math> paths possible. <math>\boxed{(\text{A})4}</math> is the correct answer.
 +
 +
==Solution 2==
 +
We can make a diagram of the roads available to Jack.
 +
<asy>
 +
size(50);
 +
defaultpen(linewidth(0.8));
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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);
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draw(D--C);
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draw(E--G);
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</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);
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draw(D--C,dotted);
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draw(E--G,dotted);
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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 <math>\boxed{A}</math>.
 +
 +
 +
 +
Note: This is not recommended in more complicated problems (e.g. Jill's house is 1000 blocks east and 400 blocks north of Jack's house).
  
 
==See Also==
 
==See Also==
 
{{AMC8 box|year=2014|num-b=10|num-a=12}}
 
{{AMC8 box|year=2014|num-b=10|num-a=12}}
 
{{MAA Notice}}
 
{{MAA Notice}}

Latest revision as of 14:29, 25 August 2025

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?

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

Insight

This is a combonatorics problem. For these problems, you should start by counting the total, then subtracting the ways that don't work. Using word rearrangements, we can find the total number of paths, then find the number of paths through the dangerous intersection using the same method. Then you subtract the first from the second. This is called Complementary Counting -JasonDaGoat

Video Solution (CREATIVE THINKING)

https://youtu.be/YCr5GFcmdcI

~Education, the Study of Everything



Video Solution

https://www.youtube.com/watch?v=rxrQLNxESW0 ~David

https://youtu.be/pWLm41JtiCw ~savannahsolver

Solution 1

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. $\boxed{(\text{A})4}$ 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 $\boxed{A}$.


Note: This is not recommended in more complicated problems (e.g. Jill's house is 1000 blocks east and 400 blocks north of Jack's house).

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

These problems are copyrighted © by the Mathematical Association of America, as part of the American Mathematics Competitions. AMC Logo.png

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