Difference between revisions of "2016 AMC 10B Problems/Problem 22"

m (Solution 2 (Cheap Solution))
m
 
Line 24: Line 24:
  
 
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 $10$ 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 1

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 $Z$ as A beat exactly 10 teams and we are choosing 2 of them. 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 2 (Cheap 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 $1/4 \cdot 1330=332.5$. Looking at the answer choices, we find that $332.5$ is closer to $385$ than any other answer choices, and that the next closest is $665$ which is twice of $332.5$, so our answer is $385$ which is $\boxed{\textbf{(A)}}$.


Speaking of cheap solutions, you can let the teams be $1,2,3,..21$ and have $i$ beat $i+1,i+2, ... i+10 \pmod{21}$ and lose to $i+11,i+12, ... i+20 \pmod{21}.$ You can choose any random team as $A$ for $21$ ways but overcount since set, so divide by $3$ so multiply by $7.$ Now we can proceed w/casework (which isn't too hard). Eventually you realize the sum is $7(1+2+\cdots+10) = \boxed{\textbf{(A)}}$.

See Also

2016 AMC 10B (ProblemsAnswer KeyResources)
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. AMC logo.png

Invalid username
Login to AoPS