A teacher has a class of 6 students and she wants to break them into three groups of 2 students. How many ways can she do that?

Guest Dec 1, 2020

#1**+1 **

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

CPhill Dec 1, 2020

#2**0 **

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.

Guest Dec 1, 2020