There are many different combinations of 3 different players (player a, player b, player c,; player a, player b, player d,; player a, player b, player e,; etc.) that could be in any given group. The question needs clarification.
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There are 5C3 = 10 groups.
if order does not matter then, (23x22x21x20x19)/(5x4x3x2x1) = 33,649
two
84 = 9 choose 3 = 9! / 3! / (9-3)!
30