Suppose that the 6 students were represented by the letters {A, B, C, D, E, F}. So:
6C2 ={A, B} | {A, C} | {A, D} | {A, E} | {A, F} | {B, C} | {B, D} | {B, E} | {B, F} | {C, D} | {C, E} | {C, F} | {D, E} | {D, F} | {E, F} (total: 15)
4C2 ={A, B} | {A, C} | {A, D} | {B, C} | {B, D} | {C, D} (total: 6) - But notice these 6 ways are all represented in the first group of 15, no matter what 4 letters you choose.
2C2 =AB, or whichever 2 letters you choose.
Therefore ( I think), [1 x 6 x 15] / 3! =15 Distinct ways of choosing the 3 groups of 2.