Difference between revisions of "2016 AMC 10B Problems/Problem 22"
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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}} |
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 |
The problems on this page are copyrighted by the Mathematical Association of America's American Mathematics Competitions.