The order in which the two players will be selected from the 8 trying players is not important. The most important thing is to have the two players.
Therefore, this is a question of combination.
The number of groups = 8C2 = 8!/[2!*(8-2)!] = (8*7*6*5*4*3*2*1)/[(2*1)*(6*5*4*3*2*1)] = 28 groups