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Difference between revisions of "2016 AMC 12B Problems/Problem 20"

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==Solution==
 
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We use complementary counting. Firstly, because each team played <math>20</math> other teams, there are <math>21</math> teams total. All sets that do not have <math>A</math> beat <math>B</math>, <math>B</math> beat <math>C</math>, and <math>C</math> beat <math>A</math> have one team that beats both the other teams. Thus we must count the number of sets of three teams such that one team beats the two other teams and subtract that number from the total number of ways to choose three teams.
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There are <math>21</math> ways to choose the team that beat the two other teams, and <math>\binom{10}{2} = 45</math> to choose two teams that the first team both beat. This is <math>21 * 45 = 945</math> sets. There are <math>\binom{21}{3} = 1330</math> sets of three teams total. Subtracting, we obtain <math>1330 - 945 = \boxed{ A) 385}</math> as our answer.
  
 
==See Also==
 
==See Also==
 
{{AMC12 box|year=2016|ab=B|num-b=19|num-a=21}}
 
{{AMC12 box|year=2016|ab=B|num-b=19|num-a=21}}
 
{{MAA Notice}}
 
{{MAA Notice}}

Revision as of 01:08, 24 February 2016

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

We use complementary counting. Firstly, because each team played $20$ other teams, there are $21$ teams total. All sets that do not have $A$ beat $B$, $B$ beat $C$, and $C$ beat $A$ have one team that beats both the other teams. Thus we must count the number of sets of three teams such that one team beats the two other teams and subtract that number from the total number of ways to choose three teams.

There are $21$ ways to choose the team that beat the two other teams, and $\binom{10}{2} = 45$ to choose two teams that the first team both beat. This is $21 * 45 = 945$ sets. There are $\binom{21}{3} = 1330$ sets of three teams total. Subtracting, we obtain $1330 - 945 = \boxed{ A) 385}$ as our answer.

See Also

2016 AMC 12B (Problems • Answer Key • Resources)
Preceded by
Problem 19 		Followed by
Problem 21
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 12 Problems and Solutions

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