Art of Problem Solving
AoPS Online
Math texts, online classes, and more
for students in grades 5-12.
Visit AoPS Online ‚
Books for Grades 5-12 Online Courses
Beast Academy
Engaging math books and online learning
for students ages 6-13.
Visit Beast Academy ‚
Books for Ages 6-13 Beast Academy Online
AoPS Academy
Small live classes for advanced math
and language arts learners in grades 2-12.
Visit AoPS Academy ‚
Find a Physical Campus Visit the Virtual Campus

2017 AIME II Problems/Problem 11

Revision as of 20:04, 27 September 2019 by Rzlng (talk | contribs) (→‎Solution 2 (complementary counting))

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 $2^{{5\choose2}}-(5 \cdot 2^6+5 \cdot 2^6-5\cdot 4 \cdot 2^3)=\boxed{544}$.

Solution 2 (complementary counting)

The only way a town does not meet the conditions in the question is if the town has either all roads leading towards it or all roads leading away from it. For example, if all roads lead away from Town $A$, there is no way to reach the town starting from Towns $B$, $C$, $D$, or $E$. If all roads lead towards Town $A$, there is no way to reach any other town starting from Town $A$. Thus, we will count the ways this occurs. (To show this, first note that there must always be a directed path that visits every town, say $A \rightarrow B \rightarrow C \rightarrow D \rightarrow E.$ (Redei's Theorem). Taking cases, it can be seen that if $A$ does not have a direct road to all other towns, then $E$ must have its roads leading into it.)


$\textbf{Case 1}$. WLOG, Town $A$ has all roads leading towards it.

In this case, the four roads leading to Town $A$ have already been determined. There are $6$ roads that have not been given directions. Each of these roads have $2$ options: it can lead towards one town or the other town. Thus, there are $2^6$ arrangements of roads. However, we must consider the $5$ towns in which this scenario can occur: there are $5 \cdot 2^6=320$ arrangements.


$\textbf{Case 2}$. WLOG, Town $A$ has all roads leading away.

Notice that this case is symmetrical to the first case. Thus, there are $320$ arrangements.


$\textbf{Case 3}$. WLOG, Town $A$ has all roads leading towards it and Town $B$ has all roads leading away.

We must check for double-counted cases. Drawing a flow diagram, we see that this case determines $7$ roads. For the undetermined $3$ roads, there are $2^3=8$ arrangements. However, we must again consider the different ways that this case can occur. There are $5$ choices for towns with roads leading towards it $4$ choices for roads leading away. Thus, the total double-counted cases are $5 \cdot 4 \cdot 8=160$ arrangements.

We must subtract the cases we counted from the total. There are a total of $10$ roads with $2$ possible arrangements each. Therefore the total number of cases is $2^{10} =1024$, and the number of cases that meet the conditions is $1024 - (320+320-160) = \boxed{544}$ arrangements.

~ jf

Solution 3

Assume the five towns are named A, B, C, D, and E. Draw roads that connect each of the five towns. First, figure out how many total roads there are with no restrictions. This would be $2^{10}$, or $1024$. Next, add together the number of restrictions (or incorrect cases) there are. One incorrect case happens if we can't get to one of the five towns. When that happens, any roads that connect that town with any of the other four towns must be one way. Four roads are invalidated (because they are one way) as, assuming we can't get to town A, roads AB, AC, AD, and AE would all be one way. Since there are ${{5\choose1}}$ ways of choosing which town to restrict, there are $5\cdot2^{10-4}$, or $320$ restrictions in this case. The next case is when we can't get to two of the towns. Let's assume we don't want to get to towns A and B. Then, roads AC, AD, AE, BC, BD, and BE are all invalidated. Since there are ${{5\choose2}}$ of choosing which two towns to restrict, there are $10\cdot2^{10-6}$, or $160$ restrictions in this case. Notice that when we restrict three towns, it is the same thing as restricting two towns. For example, if we restricted entry to towns C, D, and E from towns A and B, it's the same thing as if we restricted entry to towns A and B from towns C, D, and E, as the exact same roads are invalidated in both cases. Thus, our final answer is $1024-(320+160) = \boxed{544}$ arrangements.

Solution by IronicNinja~

See Also

Another way to state this problem is finding the number of strongly connected labeled tournaments on $5$ nodes. Finding a strongly connected labeled tournament is the same as finding a way to make all the roads one-way such that there is always a way to get from one town to another. This link gives the number of strongly connected labeled tournaments on $n$ nodes for small $n$.

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

The problems on this page are copyrighted by the Mathematical Association of America's American Mathematics Competitions. AMC logo.png

Retrieved from "https://artofproblemsolving.com/wiki/index.php?title=2017_AIME_II_Problems/Problem_11&oldid=110005"