Permutation1.1

Question

If there are 9 students in a class and each team contain 3 students then how many number of ways 9 students can be partitioned into 3 teams?



Why is this not 9C3*6C3*3C3?

Answer 1

To make #1 group having ⁹C3 ways and #2 group having ⁶C3 ways and last #3 group have 1 way to make a group with remain students.
With ordered group total ways are ⁹C3 * ⁶C3 * ³C₃.
But for unordered group  ⁹C3 * ⁶C3 * ³C₃ divided by 3!.
 

Comments

For ordered group consider that the teams are named A, B and C. For the other case assume that the team will be known by the captain's name.