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2019 Copa Libertadores de Futsal
Copa CONMEBOL Libertadores de Futsal Argentina 2019
The competition was contested by 12 teams: the title holders, one entry from each of the ten CONMEBOL associations, plus an additional entry from the host association.
The draw of the tournament was held on 4 July 2019, 13:00 ART (UTC−3), at the Predio de Ezeiza of the Argentine Football Association in Buenos Aires. The twelve teams were drawn into three groups of four. The following three teams were seeded:
Group A: the title holders
Group B: the representative (champions) from the host association (Argentina)
Group C: the representative from the association which were the runners-up from the previous edition (Brazil)
The other teams were seeded based on the results of their association in the 2018 Copa Libertadores de Futsal, with the additional entry from the host association seeded last. Each group, apart from the seeded team, contained one team from each of Pot 1, Pot 2, and Pot 3. Teams from the same association could not be drawn into the same group.
Each team has to submit a squad of 14 players, including a minimum of two goalkeepers.
Group stage
The top two teams of each group and the two best third-placed teams advance to the quarter-finals.
Tiebreakers
The teams are ranked according to points (3 points for a win, 1 point for a draw, 0 points for a loss). If tied on points, tiebreakers are applied in the following order (Regulations Article 21):
Results in head-to-head matches between tied teams (points, goal difference, goals scored);
In the quarter-finals, semi-finals and final, extra time and penalty shoot-out would be used to decide the winner if necessary (no extra time would be used in the play-offs for third to twelfth place).