We received a number of solutions for this problem with various
degrees of explanation and completeness. Some of them are below.
Well done everyone.
Adam from Bradfield School, Karina fromThe Manor Preparatory School
and Stefan from Ecclesfield School gave a solution to the first part
of the problem:
Each team wins two matches and loses two matches
Jack from Ecclesfield School offered some explanation for both parts
of the question, with a similar explanation offered by Matt of
Tapton School and Rebecca of The Chase High School:
The number of games possible has to be divisible by the number of
teams (editor`s comment: Why?). In the first league there are 10
possible games and 5 teams, so that is 2 games they have to win
each.
In the second league the number of games possible was 15 and the
number of teams was 6 so that wasn`t a whole number of games.
One solution for the first league would be:
Team 1 vs Team 2 (Team 1 won)
Team 1 vs Team 3 (team 1 won)
1 vs 4 (4 won)
1 vs 5 (5 won)
2 vs 3 (2 won)
2 vs 4 (2 won)
2 vs 5 (5 won)
3 vs 4 (3 won)
3 vs 5 (3 won)
4 vs 5 (4 won)
Matt from Tapton Comprehensive sent the folowing solution:
It is possible, for each of the five Football teams must win 2 games
with same score and lose 2 games with the same score as that with
which they won.
The reason why you could have 5 teams is because there would be 10
matches and 5 teams and 10 divides by 5 and that equals 2 games each
that each team must win.
It would be impossible because each Hockey team could either win 0
games and lose 5 games, win 1 game and lose 4 games, win 2 games and
lose 3 games, win 3 games and lose 2 games, win 4 games and lose 1
game or win all 5 games. But none of these would be even. The reason
why you could not have six teams leading because there are 15
matches and 6 teams and 15 will not divide into 6 exactly.
Finally, a more general solution offered by Andrei of School 205
Bucharest:
I observed that, if there were 5 teams, there would be 10 matches
played. This result was obtained by adding all possible
combinations:
The first team has to play with all other 4 (4 matches)
The second team also plays 4 matches, but one was already taken into
account (3 matches).
The third team plays 4 matches, but two were already taken into
account (2 matches).
The fourth team plays 4 matches, but 3 were already taken into
account (1 match).
The fifth team plays 4 matches, but all were already taken into
account (no matches).
So, the total is: 4 + 3 + 2 + 1 = 4 * 5/2 = 10 (matches).
In this case it is possible for all teams to be league leaders, as
it is possible for every team to win 2 matches, and to lose 2:
10/5 = 2 (won matches).
Now, if there are six teams, the total number of played matches
would be: 5 + 4 + 3 + 2 + 1 = 5 * 6/2 = 15 (matches).
As 15 is not divisible by 6, it`s impossible that every team will
win the same number of matches. So, it`s impossible for all the
teams to be league leaders.
Now, I look more generally at the problem:
If the number of teams is even (2k), than the total number of
matches would be:
(2k - 1) + (2k - 2) + ... + 1 =
2k(2k - 1)2
= k(2k -1)
This result should be divided by the number of teams:
k(2k - 1)2k
=
(2k - 1)2
This number will never be divisible by 2, so, if there is an even
number of teams, not all of them could be league leaders.
Now, I analyze the case of an odd number of teams (2k + 1); the
total number of matches is:
2k + (2k - 1) + (2k - 2) + ... + 1 =
(2k + 1)2k2
= k(2k+ 1)
This number should be divided by the number of teams (2k+1):
k(2k + 1)2k + 1
= k
So, if the number of teams is odd, then all of them could be league
leaders.
Note: The total number of matches played in a round robin tournament
with p teams, each playing only once with all others is the number
of ways of selecting a group of 2 teams from a group of p, when
order does not matter, i.e. combinations: