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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>.
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==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*1330=332.5</math>. Looking at the answer choices, we find that 385 is closer to 332.5 than any other answer choices, so our answer is <math>385</math> which is <math>\boxed{\textbf{(A)}}</math>.
  
 
==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}}

Revision as of 19:01, 29 May 2019

Problem

A set of teams held a round-robin tournament in which every team played every other team exactly once. Every team won $10$ games and lost $10$ games; there were no ties. How many sets of three teams $\{A, B, C\}$ were there in which $A$ beat $B$, $B$ beat $C$, and $C$ beat $A?$

$\textbf{(A)}\ 385 \qquad \textbf{(B)}\ 665 \qquad \textbf{(C)}\ 945 \qquad \textbf{(D)}\ 1140 \qquad \textbf{(E)}\ 1330$

Solution

There are $21$ teams. Any of the $\tbinom{21}3=1330$ sets of three teams must either be a fork (in which one team beat both the others) or a cycle:

[asy]size(7cm);label("X",(5,5));label("Z",(10,0));label("Y",(0,0));draw((4,4)--(1,1),EndArrow);draw((6,4)--(9,1),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] But we know that every team beat exactly $10$ other teams, so for each possible $X$ at the head of a fork, there are always exactly $\tbinom{10}2$ choices for $Y$ and $Z$. Therefore there are $21\cdot\tbinom{10}2=945$ forks, and all the rest must be cycles.

Thus the answer is $1330-945=385$ which is $\boxed{\textbf{(A)}}$.

Solution

Since there are $21$ teams and for each set of three teams there is a cycle, there are a total of $\tbinom{21}3=1330$ cycles of three teams. Because about $1/4$ of the cycles $\{A, B, C\}$ satisfy the conditions of the problems, our answer is close to $1/4*1330=332.5$. Looking at the answer choices, we find that 385 is closer to 332.5 than any other answer choices, so our answer is $385$ which is $\boxed{\textbf{(A)}}$.

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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