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Question

An international curling competition has 10 teams. in the round-robin portion of the competition. 1)determine the number of games that are played during the round-robin portion of the competition. 2)in the medal round, there are two semifinal games. in one, the first place team plays the fourth-place team, and in the other, the second-place team plays the third-place team. determine the number of different possible pairings for the semifinal games.explain your answer.

Answer 1

In a round-robin tournament, each team plays 1 game with each of the other teams 10 teams.
Number of games played = 10!/(10-2)!
= 10!/8!
= 90

Answer 2

1v2
1v3
1v4
...
1v10 (9)
2v3 (we skip 2v1 because that's the same game as 1v2)
2v4
...
2v10 (8)
3v4
3v5
...
3v10 (7)
...

Total number of games =
9+8+7+...+2+1= 45

1v5 is the same game as 5v1; D.W.'s method counts them as distinct games. If the round robin has each team act as home team in one game and "away" team in the other game, then 1v5 is different than 5v1 and D.W. is correct.

The first-place team can be any of 10
The second-place can be any of the remaining 9
The third-place team can be any of the remaining 8
The fourth-place-team can be any of the remaining 7

Total number of possibilities is 10*9*8*7 if it makes a difference which is second and which is third (e.g., if the second-place team plays as "home" team instead of away.

If not, we have to reduce the number accordingly.

Once you have established that the ranking is A, B, C, D, then this is the same as if the ranking had been D, C, B, A or D, B, C, A or A, C, B, D.

So the number of possible pairings is (10*9*8*7)/4 = 1260.

Answer 3

Each team plays each of the other teams: 10 x 9 = 90 games. In the semi-final, for team #1 there are 10 possibilities; for team #2, there are 9 possibilities, for team #3 there are 8 possibilities, and for team #4 there are 7 possibilities. So you would take 10x7=70 for the #1-#4 game, and 9x8=72 for the #2-#3 game, a total of 142 possible pairings.