Difference between revisions of "2016 AMC 10B Problems/Problem 22"
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Thus the answer is <math>1330-945=385</math> which is <math>\boxed{\textbf{(A)}}</math>. | Thus the answer is <math>1330-945=385</math> which is <math>\boxed{\textbf{(A)}}</math>. | ||
− | ==Solution== | + | ==Solution 2== |
Since there are <math>21</math> teams and for each set of three teams there is a cycle, there are a total of <math>\tbinom{21}3=1330</math> cycles of three teams. Because about <math>1/4</math> of the cycles <math>\{A, B, C\}</math> satisfy the conditions of the problems, our answer is close to <math>1/4*1330=332.5</math>. Looking at the answer choices, we find that <math>332.5</math> is closer to <math>385</math> than any other answer choices, so our answer is <math>385</math> which is <math>\boxed{\textbf{(A)}}</math>. | Since there are <math>21</math> teams and for each set of three teams there is a cycle, there are a total of <math>\tbinom{21}3=1330</math> cycles of three teams. Because about <math>1/4</math> of the cycles <math>\{A, B, C\}</math> satisfy the conditions of the problems, our answer is close to <math>1/4*1330=332.5</math>. Looking at the answer choices, we find that <math>332.5</math> is closer to <math>385</math> than any other answer choices, so our answer is <math>385</math> which is <math>\boxed{\textbf{(A)}}</math>. | ||
Revision as of 21:28, 29 May 2019
Contents
Problem
A set of teams held a round-robin tournament in which every team played every other team exactly once. Every team won games and lost games; there were no ties. How many sets of three teams were there in which beat , beat , and beat
Solution
There are teams. Any of the sets of three teams must either be a fork (in which one team beat both the others) or a cycle:
But we know that every team beat exactly other teams, so for each possible at the head of a fork, there are always exactly choices for and . Therefore there are forks, and all the rest must be cycles.
Thus the answer is which is .
Solution 2
Since there are teams and for each set of three teams there is a cycle, there are a total of cycles of three teams. Because about of the cycles satisfy the conditions of the problems, our answer is close to . Looking at the answer choices, we find that is closer to than any other answer choices, so our answer is which is .
See Also
2016 AMC 10B (Problems • Answer Key • Resources) | ||
Preceded by Problem 21 |
Followed by Problem 23 | |
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 AMC 10 Problems and Solutions |
The problems on this page are copyrighted by the Mathematical Association of America's American Mathematics Competitions.