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Difference between revisions of "2017 AIME II Problems/Problem 11"
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There are two towns not on the loop; call them Town <math>A</math> and Town <math>B</math>. Without loss of generality assume <math>A</math> leads to <math>B</math>. Because a road must lead to <math>A</math>, the town where this road comes from must be on the loop. Therefore <math>A</math> and therefore <math>B</math> can be reached from the loop. Because a road must lead out of <math>B</math>, the town it leads to must be on the loop. Therefore the loop can be reached from <math>B</math> and also <math>A</math>.
 
There are two towns not on the loop; call them Town <math>A</math> and Town <math>B</math>. Without loss of generality assume <math>A</math> leads to <math>B</math>. Because a road must lead to <math>A</math>, the town where this road comes from must be on the loop. Therefore <math>A</math> and therefore <math>B</math> can be reached from the loop. Because a road must lead out of <math>B</math>, the town it leads to must be on the loop. Therefore the loop can be reached from <math>B</math> and also <math>A</math>.
  
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The number of good configurations is the total number of configurations minus the number of bad configurations. There are <math>2^{{5\choose2}}</math> total configurations. To find the number of bad configurations in which a town exists such that all roads lead to it, there are <math>5</math> ways to choose this town and <math>2^6</math> ways to assign the six other roads that do not connect to this town. The same logic is used to find the number of bad configurations in which a town exists such that all roads lead out of it. It might be tempting to conclude that there are <math>5\times2^6+5\times2^6</math> bad configurations, but the configurations in which there exists a town such that all roads lead to it and a town such that all roads lead out of it are overcounted. There are <math>5</math> ways to choose the town for which all roads lead to it, <math>4</math> ways to choose the town for which all roads lead out of it, and <math>2^3</math> ways to assign the remaining <math>3</math> roads not connected to either of these towns. Therefore, the answer is <math>2^{{5\choose2}}-(5\times2^6+5\times2^6-5\times4\times2^3)=\boxed{544}</math>.
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The number of good configurations is the total number of configurations minus the number of bad configurations. There are <math>2^{{5\choose2}}</math> total configurations. To find the number of bad configurations in which a town exists such that all roads lead to it, there are <math>5</math> ways to choose this town and <math>2^6</math> ways to assign the six other roads that do not connect to this town. The same logic is used to find the number of bad configurations in which a town exists such that all roads lead out of it. It might be tempting to conclude that there are <math>5 \cdot 2^6+5 \cdot 2^6</math> bad configurations, but the configurations in which there exists a town such that all roads lead to it and a town such that all roads lead out of it are overcounted. There are <math>5</math> ways to choose the town for which all roads lead to it, <math>4</math> ways to choose the town for which all roads lead out of it, and <math>2^3</math> ways to assign the remaining <math>3</math> roads not connected to either of these towns. Therefore, the answer is <math>2^{{5\choose2}}-(5 \cdot 2^6+5 \cdot 2^6-5\cdot 4 \cdot 2^3)=\boxed{544}</math>.
  
 
=See Also=
 
=See Also=
 
{{AIME box|year=2017|n=II|num-b=10|num-a=12}}
 
{{AIME box|year=2017|n=II|num-b=10|num-a=12}}
 
{{MAA Notice}}
 
{{MAA Notice}}

Revision as of 17:38, 23 March 2017

Problem

Five towns are connected by a system of roads. There is exactly one road connecting each pair of towns. Find the number of ways there are to make all the roads one-way in such a way that it is still possible to get from any town to any other town using the roads (possibly passing through other towns on the way).

Solution

It is obvious that any configuration of one-way roads which contains a town whose roads all lead into it or lead out of it cannot satisfy the given. We claim that any configuration which does not have a town whose roads all lead into it or lead out of it does satisfy the given conditions. Now we show that a loop of $3$ or more towns exist. Pick a town, then choose a neighboring town to travel to $5$ times. Of these $6$ towns visited, at least two of them must be the same; therefore there must exist a loop of $3$ or more towns because a loop of $2$ towns cannot exist. We want to show that the loop can be reached from any town, and any town can be reached from the loop.

$\textbf{Case 1}$. The loop has $5$ towns. Clearly every town can be reached by going around the loop.

$\textbf{Case 2}$. The loop has $4$ towns. The town not on the loop must have a road leading to it. This road comes from a town on the loop. Therefore this town can be reached from the loop. This town not on the loop must also have a road leading out of it. This road leads to a town on the loop. Therefore the loop can be reached from the town.

$\textbf{Case 3}$. The loop has $3$ towns. There are two towns not on the loop; call them Town $A$ and Town $B$. Without loss of generality assume $A$ leads to $B$. Because a road must lead to $A$, the town where this road comes from must be on the loop. Therefore $A$ and therefore $B$ can be reached from the loop. Because a road must lead out of $B$, the town it leads to must be on the loop. Therefore the loop can be reached from $B$ and also $A$.

The number of good configurations is the total number of configurations minus the number of bad configurations. There are $2^{{5\choose2}}$ total configurations. To find the number of bad configurations in which a town exists such that all roads lead to it, there are $5$ ways to choose this town and $2^6$ ways to assign the six other roads that do not connect to this town. The same logic is used to find the number of bad configurations in which a town exists such that all roads lead out of it. It might be tempting to conclude that there are $5 \cdot 2^6+5 \cdot 2^6$ bad configurations, but the configurations in which there exists a town such that all roads lead to it and a town such that all roads lead out of it are overcounted. There are $5$ ways to choose the town for which all roads lead to it, $4$ ways to choose the town for which all roads lead out of it, and $2^3$ ways to assign the remaining $3$ roads not connected to either of these towns. Therefore, the answer is $2^{{5\choose2}}-(5 \cdot 2^6+5 \cdot 2^6-5\cdot 4 \cdot 2^3)=\boxed{544}$.

See Also

2017 AIME II (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
All AIME Problems and Solutions

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