Difference between revisions of "2006 Canadian MO Problems/Problem 4"
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| + | (b) By using complementary counting, it is not hard to find that the answer is <math>\dfrac{n(n+1)(2n+1)}{6}</math>. | ||
Revision as of 18:55, 16 December 2012
Problem
Consider a round robin tournament with 
teams, where two teams play exactly one match and there are no ties. We say that the teams 
, 
, and 
form a cycle triplet if beats , beats , and beats .
(a) Find the minimum number of cycle triplets possible.
(b) Find the maximum number of cycle triplets possible.
Solution
This problem needs a solution. If you have a solution for it, please help us out by adding it.
See also
| 2006 Canadian MO (Problems) | ||
| Preceded by Problem 3 |
1 • 2 • 3 • 4 • 5 | Followed by Problem 5 |
(a) Clearly the answer is 0.
(b) By using complementary counting, it is not hard to find that the answer is 
.
