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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?

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    @Joffan "if every team plays each other *twice*" – 2017-01-29
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    @Joffan I guess I have no idea what a fixture is; I just assumed it was another word for "match". – 2017-01-29
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    Well, that's what I get for thinking I can math goodly. – 2017-01-29
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    @Nij perhaps you understand my original point then :-) – 2017-01-29

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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.

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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.