Difference between revisions of "2023 AMC 8 Problems/Problem 8"
(→Problem) |
|||
| Line 5: | Line 5: | ||
<cmath>\begin{tabular}{c | c} Player & Result \\ \hline Lola & \texttt{111011}\\ Lolo & \texttt{101010}\\ Tiya & \texttt{010100}\\ Tiyo & \texttt{??????} \end{tabular}</cmath> | <cmath>\begin{tabular}{c | c} Player & Result \\ \hline Lola & \texttt{111011}\\ Lolo & \texttt{101010}\\ Tiya & \texttt{010100}\\ Tiyo & \texttt{??????} \end{tabular}</cmath> | ||
<math>\textbf{(A)}\ \texttt{000101} \qquad \textbf{(B)}\ \texttt{001001} \qquad \textbf{(C)}\ \texttt{010000} \qquad \textbf{(D)}\ \texttt{010101} \qquad \textbf{(E)}\ \texttt{011000}</math> | <math>\textbf{(A)}\ \texttt{000101} \qquad \textbf{(B)}\ \texttt{001001} \qquad \textbf{(C)}\ \texttt{010000} \qquad \textbf{(D)}\ \texttt{010101} \qquad \textbf{(E)}\ \texttt{011000}</math> | ||
| + | |||
| + | == 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. | ||
| + | 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> | ||
Revision as of 19:50, 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 
and 
represent a win or loss, respectively. For example, Lola won five matches and lost the fourth match. What was Tiyo’s win-loss record?
![\[\begin{tabular}{c | c} Player & Result \\ \hline Lola & \texttt{111011}\\ Lolo & \texttt{101010}\\ Tiya & \texttt{010100}\\ Tiyo & \texttt{??????} \end{tabular}\]](http://latex.artofproblemsolving.com/c/1/0/c108a299c4eace651f02848c15ca7a646f6adff0.png)
Solution
In total, there will be 
games because there are 
ways to choose a pair of people from the four players and each player will play each other player exactly twice. Each of these 
games will have winner and loser, so there will be a total of 's and 's in the win-loss table. Therefore, Tiyo will have 
's and 
's in his record.
Now, all we have to do is figure out the order of these 's and '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 
winners and losers which means that every column of the win-loss table should have 's and 's. Looking at the filled-in table so far, we see that columns 
and 
need one more , so Tiyo must have 's in those columns and 's go in the others.
Therefore, our answer is 
