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Difference between revisions of "2023 AMC 8 Problems/Problem 8"

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== Solution ==
 
== Solution ==
 
In total, there will be <math>\binom{4}{2} \cdot 2 = 6 \cdot 2 = 12</math> games because there are <math>\binom{4}{2}</math> ways to choose a pair of people from the four players and each player will play each other player exactly twice. Each of these <math>12</math> games will have <math>1</math> winner and <math>1</math> loser, so there will be a total of <math>12</math> <math>1</math>'s and <math>12</math> <math>0</math>'s in the win-loss table. Therefore, Tiyo will have <math>12-10=2</math> <math>1</math>'s and <math>12-8=4</math> <math>0</math>'s in his record.
 
In total, there will be <math>\binom{4}{2} \cdot 2 = 6 \cdot 2 = 12</math> games because there are <math>\binom{4}{2}</math> ways to choose a pair of people from the four players and each player will play each other player exactly twice. Each of these <math>12</math> games will have <math>1</math> winner and <math>1</math> loser, so there will be a total of <math>12</math> <math>1</math>'s and <math>12</math> <math>0</math>'s in the win-loss table. Therefore, Tiyo will have <math>12-10=2</math> <math>1</math>'s and <math>12-8=4</math> <math>0</math>'s in his record.
 +
 
Now, all we have to do is figure out the order of these <math>1</math>'s and <math>0</math>'s. Clearly, in every round, there are two games; the players are split into two pairs and the people in those pairs play each other. Thus, every round should have <math>2</math> winners and <math>2</math> losers which means that every column of the win-loss table should have <math>2</math> <math>1</math>'s and <math>2</math> <math>0</math>'s. Looking at the filled-in table so far, we see that columns <math>4</math> and <math>6</math> need one more <math>1</math>, so Tiyo must have <math>1</math>'s in those columns and <math>0</math>'s go in the others.  
 
Now, all we have to do is figure out the order of these <math>1</math>'s and <math>0</math>'s. Clearly, in every round, there are two games; the players are split into two pairs and the people in those pairs play each other. Thus, every round should have <math>2</math> winners and <math>2</math> losers which means that every column of the win-loss table should have <math>2</math> <math>1</math>'s and <math>2</math> <math>0</math>'s. Looking at the filled-in table so far, we see that columns <math>4</math> and <math>6</math> need one more <math>1</math>, so Tiyo must have <math>1</math>'s in those columns and <math>0</math>'s go in the others.  
 
Therefore, our answer is <math>\boxed{\textbf{(A)}\ \texttt{000101}}.</math>
 
Therefore, our answer is <math>\boxed{\textbf{(A)}\ \texttt{000101}}.</math>
  
 
~lpieleanu
 
~lpieleanu

Revision as of 20:52, 24 January 2023

Problem

Lola, Lolo, Tiya, and Tiyo participated in a ping pong tournament. Each player competed against each of the other three players exactly twice. Shown below are the win-loss records for the players. The numbers $1$ and $0$ represent a win or loss, respectively. For example, Lola won five matches and lost the fourth match. What was Tiyo’s win-loss record?

\[\begin{tabular}{c | c} Player & Result \\ \hline Lola & \texttt{111011}\\ Lolo & \texttt{101010}\\ Tiya & \texttt{010100}\\ Tiyo & \texttt{??????} \end{tabular}\] $\textbf{(A)}\ \texttt{000101} \qquad \textbf{(B)}\ \texttt{001001} \qquad \textbf{(C)}\ \texttt{010000} \qquad \textbf{(D)}\ \texttt{010101} \qquad \textbf{(E)}\ \texttt{011000}$


Solution

We can calculate the total number of wins (1's) by seeing how many matches were playes, which are 12 matches played. Then we can calculate the # of wins already on the table, which is 5 + 3 + 2 = 10 so there are 12 - 10 = 2 wins left in the mystery player. Now we will make the key observation that there are only 2 1's per column as there are 2 winners and 2 losers in each round. Strategically looking through the colums counting the 1's and putting our own 2 1's when the column isn't already full yields $\boxed{\textbf{(A)}\ \texttt{000101}}$

~SohumUttamchandani

Solution

In total, there will be $\binom{4}{2} \cdot 2 = 6 \cdot 2 = 12$ games because there are $\binom{4}{2}$ ways to choose a pair of people from the four players and each player will play each other player exactly twice. Each of these $12$ games will have $1$ winner and $1$ loser, so there will be a total of $12$ $1$'s and $12$ $0$'s in the win-loss table. Therefore, Tiyo will have $12-10=2$ $1$'s and $12-8=4$ $0$'s in his record.

Now, all we have to do is figure out the order of these $1$'s and $0$'s. Clearly, in every round, there are two games; the players are split into two pairs and the people in those pairs play each other. Thus, every round should have $2$ winners and $2$ losers which means that every column of the win-loss table should have $2$ $1$'s and $2$ $0$'s. Looking at the filled-in table so far, we see that columns $4$ and $6$ need one more $1$, so Tiyo must have $1$'s in those columns and $0$'s go in the others. Therefore, our answer is $\boxed{\textbf{(A)}\ \texttt{000101}}.$

~lpieleanu

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