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2013 AMC 12B Problems/Problem 12: Difference between revisions

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==Solution==
==Solution==
Note that cities <math>C</math> and <math>E</math> can be removed when counting paths because if a path goes in to <math>C</math> or <math>E</math>, there is only one possible path to take out of cities <math>C</math> or <math>E</math>.
So the diagram is as follows:
<asy>
unitsize(10mm);
defaultpen(linewidth(1.2pt)+fontsize(10pt));
dotfactor=4;
pair A=(1,0), B=(4.24,0), C=(5.24,3.08), D=(2.62,4.98), E=(0,3.08);
dot (A);
dot (B);
dot (D);
label("$A$",A,S);
label("$B$",B,SE);
label("$D$",D,N);
draw(A--B..D..cycle);
draw(A--D);
draw(B--D);</asy>
Now we proceed with casework. Remember that there are two ways to travel from <math>A</math> to <math>D</math>, <math>D</math> to <math>A</math>, <math>B</math> to <math>D</math> and <math>D</math> to <math>B</math>.:
Case 1 <math>A \Rightarrow D</math>: From <math>D</math>, if the path returns to <math>A</math>, then the next path must go to <math>B\Rightarrow D \Rightarrow B</math>. There are <math>2 \cdot 1 \cdot 2 = 4</math> possibilities of the path <math>ADABDB</math>. If the path goes to <math>B</math> from <math>D</math>, then the path must continue with either <math>BDAB</math> or <math>BADB</math>. There are <math>2 \cdot 2 \cdot 2 = 8</math> possibilities. So, this case gives <math>4+8=12</math> different possibilities.
Case 2 <math>A \Rightarrow B</math>: The path must continue with <math>BDADB</math>. There are <math>2 \cdot 2 = 4</math> possibilities for this case.
Putting the two cases together gives <math>12+4 = \boxed{\textbf{(D)}16}</math>


== See also ==
== See also ==
{{AMC12 box|year=2013|ab=B|num-b=11|num-a=13}}
{{AMC12 box|year=2013|ab=B|num-b=11|num-a=13}}

Revision as of 20:14, 22 February 2013

Problem

Cities $A$, $B$, $C$, $D$, and $E$ are connected by roads $AB$, $AD$, $AE$, $BC$, $BD$, $CD$, and $DE$. How many different routes are there from $A$ to $B$ that use each road exactly once? (Such a route will necessarily visit some cities more than once.) [asy] unitsize(10mm); defaultpen(linewidth(1.2pt)+fontsize(10pt)); dotfactor=4; pair A=(1,0), B=(4.24,0), C=(5.24,3.08), D=(2.62,4.98), E=(0,3.08); dot (A); dot (B); dot (C); dot (D); dot (E); label("$A$",A,S); label("$B$",B,SE); label("$C$",C,E); label("$D$",D,N); label("$E$",E,W); draw(A--B--C--D--E--cycle); draw(A--D); draw(B--D);[/asy]

$\textbf{(A)}\ 7 \qquad \textbf{(B)}\ 9 \qquad \textbf{(C)}\ 12 \qquad \textbf{(D)}\ 16 \qquad \textbf{(E)}\ 18$

Solution

Note that cities $C$ and $E$ can be removed when counting paths because if a path goes in to $C$ or $E$, there is only one possible path to take out of cities $C$ or $E$. So the diagram is as follows: [asy] unitsize(10mm); defaultpen(linewidth(1.2pt)+fontsize(10pt)); dotfactor=4; pair A=(1,0), B=(4.24,0), C=(5.24,3.08), D=(2.62,4.98), E=(0,3.08); dot (A); dot (B); dot (D); label("$A$",A,S); label("$B$",B,SE); label("$D$",D,N); draw(A--B..D..cycle); draw(A--D); draw(B--D);[/asy]

Now we proceed with casework. Remember that there are two ways to travel from $A$ to $D$, $D$ to $A$, $B$ to $D$ and $D$ to $B$.:

Case 1 $A \Rightarrow D$: From $D$, if the path returns to $A$, then the next path must go to $B\Rightarrow D \Rightarrow B$. There are $2 \cdot 1 \cdot 2 = 4$ possibilities of the path $ADABDB$. If the path goes to $B$ from $D$, then the path must continue with either $BDAB$ or $BADB$. There are $2 \cdot 2 \cdot 2 = 8$ possibilities. So, this case gives $4+8=12$ different possibilities.

Case 2 $A \Rightarrow B$: The path must continue with $BDADB$. There are $2 \cdot 2 = 4$ possibilities for this case.

Putting the two cases together gives $12+4 = \boxed{\textbf{(D)}16}$

See also

2013 AMC 12B (Problems • Answer Key • Resources)
Preceded by
Problem 11 		Followed by
Problem 13
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 AMC 12 Problems and Solutions
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