A teacher has a class of 6 students

Choose  any 2 of the 6 to be in the first  group = C(6,2)  = 15 ways to do this

Choose  any  2 of the remaining 4 to be  in the next group  = C(4,2) = 6 ways to do this

The last gtoup  of 2   will be by default

So

15 * 6   =    90 ways to do this

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.