Difference between revisions of "2016 AMC 10B Problems/Problem 22"
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label("X",(20,5));label("Z",(25,0));label("Y",(15,0));draw((19,4)--(16,1),EndArrow);draw((16,0)--(24,0),EndArrow);draw((24,1)--(21,4),EndArrow); | label("X",(20,5));label("Z",(25,0));label("Y",(15,0));draw((19,4)--(16,1),EndArrow);draw((16,0)--(24,0),EndArrow);draw((24,1)--(21,4),EndArrow); | ||
</asy> | </asy> | ||
− | But we know that every team beat exactly <math>10</math> other teams, so for each possible <math>X</math> at the head of a fork, there are always exactly <math>\tbinom{10}2</math> choices for <math>Y</math> and <math>Z</math>. Therefore there are <math>21\cdot\tbinom{10}2=945</math> forks, and all the rest must be cycles. | + | But we know that every team beat exactly <math>10</math> other teams, so for each possible <math>X</math> at the head of a fork, there are always exactly <math>\tbinom{10}2</math> choices for <math>Y</math> and <math>Z</math> as A beat exactly 10 teams and we are choosing 2 of them. Therefore there are <math>21\cdot\tbinom{10}2=945</math> forks, and all the rest must be cycles. |
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 2 (Cheap Solution)== | ==Solution 2 (Cheap Solution)== | ||
− | 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 \cdot 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 and that the next closest is<math>665</math> which is twice <math>332.5</math>, 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 \cdot 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, and that the next closest is <math>665</math> which is twice of <math>332.5</math>, so our answer is <math>385</math> which is <math>\boxed{\textbf{(A)}}</math>. |
Speaking of cheap solutions, you can let the teams be <math>1,2,3,..21</math> and have <math>i</math> beat <math>i+1,i+2, ... i+10 \pmod{21}</math> and lose to <math>i+11,i+12, ... i+20 \pmod{21}.</math> You can choose any random team as <math>A</math> for <math>21</math> ways but overcount since set, so divide by <math>3</math> so multiply by <math>7.</math> Now we can proceed w/casework (which isn't too hard). Eventually you realize the sum is <math>7(1+2+\cdots+10) = \boxed{\textbf{(A)}}</math>. | Speaking of cheap solutions, you can let the teams be <math>1,2,3,..21</math> and have <math>i</math> beat <math>i+1,i+2, ... i+10 \pmod{21}</math> and lose to <math>i+11,i+12, ... i+20 \pmod{21}.</math> You can choose any random team as <math>A</math> for <math>21</math> ways but overcount since set, so divide by <math>3</math> so multiply by <math>7.</math> Now we can proceed w/casework (which isn't too hard). Eventually you realize the sum is <math>7(1+2+\cdots+10) = \boxed{\textbf{(A)}}</math>. | ||
− | + | [[Category:Introductory_Combinatorics_Problems]] | |
==See Also== | ==See Also== | ||
{{AMC10 box|year=2016|ab=B|num-b=21|num-a=23}} | {{AMC10 box|year=2016|ab=B|num-b=21|num-a=23}} | ||
{{MAA Notice}} | {{MAA Notice}} |
Latest revision as of 17:27, 6 April 2021
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 1
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 as A beat exactly 10 teams and we are choosing 2 of them. Therefore there are forks, and all the rest must be cycles.
Thus the answer is which is .
Solution 2 (Cheap Solution)
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, and that the next closest is which is twice of , so our answer is which is .
Speaking of cheap solutions, you can let the teams be and have beat and lose to You can choose any random team as for ways but overcount since set, so divide by so multiply by Now we can proceed w/casework (which isn't too hard). Eventually you realize the sum 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 |
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