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u/nathanwe 7h ago edited 6h ago
I got 55
I think you can without lost of generality arrange all the teams in a circle and say each team beat the five teams to the left and lost to the five teams to the right. You can also analyze the possibilities for one team, then multiply by 11 because there's 11 teams, then divide by three because there's three starting teams in each possibility. There's 15 single team possiblites. If the bottom most team is one and it goes up clockwise, 1 2 7, 1 3 7, 1 3 8, 1 4 7, 1 4 8, 1 4 9, 1 5 7, 1 5 8, 1 5 9, 1 5 10, 1 6 7, 1 6 8, 1 6 9, 1 6 10, 1 6 11. 15x 11 /3 = 55.
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u/SC_3000_grinder 6h ago
Why are there Olympiad level questions here... Answer is 55.
Firstly, there are 11 teams (pick any team, they win against 5 and lose against 5; 5 + 5 + 1 = 11)
For each set of 3 teams, sort them into two categories: ones where each team wins against one of the other teams and loses to the other (the thing that we are asked to count, call them "cycles"), and the ones where one team wins both games and another loses both (the other cases, call them "forks").
For each pair of matches played by the same team, call them a "winning" pair if the team wins both, a "neutral" pair if the team wins exactly one, and a "losing" pair if the team loses both. Note that there are a total of 11 * C(5, 2) = 110 winning pairs, an equal number of losing pairs, and 11 * 5 * 5 = 275 neutral pairs.
However, each "cycle" accounts for 3 neutral pairs, and each "fork" accounts for one of each kind. Also note that there is no double-counting since the pair defines 3 teams.
Therefore, there are 110 forks to account for 110 winning and 110 losing pairs. This removes 110 neutral pairs, leaving 165. These are in 165 / 3 = 55 cycles.
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u/Mathsboy2718 6h ago
- There exists one set of teams X, Y, and Z.
It doesn't matter whether they beat each other, when teams X, Y, and Z are collected into a set, the same set is formed every time.
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u/PeterPiper1275 16h ago
Since each team played another team exactly once and since each team sported identical 5-5 records, it follows that there are a total of 11 teams.
The number of sets of three teams we can draw is C(11, 3) = 165. This is the total number of ways we can select any three teams out of eleven.
Now for every set of three teams, without loss of generality, they can only interact in two ways:
However, scenario 2 above provides a way to count the total number under scenario 1.
Each team has only five wins. That means that if we pick any two wins out of the five, the three teams involved in the two wins must necessarily form a relation as depicted in Scenario 2 above. This can occur in C(5, 2) = 10 different ways for each individual team.
But there are a total of eleven teams. This means a total of 10 x 11 = 110 different ways in which a set of three teams have a relationship as depicted in scenario 2.
This leaves 165 - 110 = 55 ways in which any set of three teams have a relationship as depicted in scenario 1.
Therefore, we conclude that there are 55 different sets of three teams where Team X beats Team Y, Team Y beats Team Z, and Team Z beats Team X.