Difference between revisions of "2004 AMC 8 Problems/Problem 5"
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== Problem == | == Problem == | ||
| − | The losing team of each game is eliminated from the tournament. If sixteen teams compete, how many games will be played to determine the winner? | + | Ms. Hamilton's eighth-grade class wants to participate in the annual three-person-team basketball tournament. The losing team of each game is eliminated from the tournament. If sixteen teams compete, how many games will be played to determine the winner? |
<math> \textbf{(A)}\ 4\qquad\textbf{(B)}\ 7\qquad\textbf{(C)}\ 8\qquad\textbf{(D)}\ 15\qquad\textbf{(E)}\ 16 </math> | <math> \textbf{(A)}\ 4\qquad\textbf{(B)}\ 7\qquad\textbf{(C)}\ 8\qquad\textbf{(D)}\ 15\qquad\textbf{(E)}\ 16 </math> | ||
| − | =Solution 1= | + | ==Solution 1== |
| − | + | Note that the winning team will the be the only team that wins all of the games. Therefore, to find the total number of games to determine the winner has a 1:1 correspondence to the number of ways to determine the losers. ( Think of it this way: If you want to select two balls from a bag of 6 balls, it is analogous to selecting the 4 balls that you don't want to select, both are 6 choose 2.). There are 15 losing teams, and since each round is unique, there are 15 total rounds. | |
| + | |||
| + | ~Brackie1331 | ||
==Solution 2== | ==Solution 2== | ||
Latest revision as of 09:37, 16 May 2024
Contents
Problem
Ms. Hamilton's eighth-grade class wants to participate in the annual three-person-team basketball tournament. The losing team of each game is eliminated from the tournament. If sixteen teams compete, how many games will be played to determine the winner?
Solution 1
Note that the winning team will the be the only team that wins all of the games. Therefore, to find the total number of games to determine the winner has a 1:1 correspondence to the number of ways to determine the losers. ( Think of it this way: If you want to select two balls from a bag of 6 balls, it is analogous to selecting the 4 balls that you don't want to select, both are 6 choose 2.). There are 15 losing teams, and since each round is unique, there are 15 total rounds.
~Brackie1331
Solution 2
There will be 
games the first round, 
games the second round, 
games the third round, and 
game in the final round, giving us a total of 
games. 
.
See Also
| 2004 AMC 8 (Problems • Answer Key • Resources) | ||
| Preceded by Problem 4 |
Followed by Problem 6 | |
| 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 AJHSME/AMC 8 Problems and Solutions | ||
The problems on this page are copyrighted by the Mathematical Association of America's American Mathematics Competitions. 
