Let the 4 schools be A, B , C and D
A plays B, C and D
B plays C, D (they have already played A)
C plays D ( they have already played A and B
D has already played A, B and C
The players from A play a total of
(4 players from A ) ( 4 players* 3 schools) (3 games each) = 144 games
The players from B play a total of
(4 players from B) (4 players *2 schools)(3 games each ) = 96 games
The players from C play a total of
(4 players from C) ( 4 players * 1 school ) (3 games each) = 48 games
144 + 96 + 48 =
48 ( 3 + 2 + 1) =
48 (6) =
288 games