If there are $20$ teams in a league, the total number of fixtures are $380$ if every team plays each other twice.
How would you come to that number where there are any number of teams?
$\endgroup$ 43 Answers
$\begingroup$If there are $n$ teams, a round robin consists of $\frac 12n(n-1)$ games. You choose one team out of $n$, then choose the other out of the remaining $n-1$ but have counted each game twice because there are two orders to choose the teams.
$\endgroup$ $\begingroup$OK, since we now have the correct count of fixtures for the opening question...
$20$ teams in a league playing two matches against each opponent. Each team then is scheduled to play $19\times 2=38$ matches, which gives a total of $38*20/2=380$ matches in total, since each match involves two teams.
Thus for a league of $n$ teams, each team playing two matches against each opponent;each team will play $2(n-1)$ matches and in total there will be $n\cdot 2(n-1)/2 = $$\fbox{$n(n-1)$}$ matches.
$\endgroup$ $\begingroup$I will use combination. To know the amount of fixtures without taking home and away into consideration, it would be $~^{20}C_2 =190~$. Multiply this by $~2~$ because of home and away. This is then $~380~$. Thus if there are $~n~$ teams, no of matches is $~^nC_2 \cdot 2~$.
Thank you.
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