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 2== | + | ==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*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 03:12, 31 December 2019
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:
![[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]](http://latex.artofproblemsolving.com/c/f/b/cfb866b5865b62be078af1b1915ffc5fed69bab6.png)
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 (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, 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. 
