How many ways are there to put 6 balls in 3 boxes if the balls are distinguishable but the boxes are not?

ProffesorNobody Apr 16, 2019

#1**+1 **

We break this up into cases. To create 6, with three numbers, we can use: 6-0-0, 5-1-0, 4-2-0, 4-1-1, 3-3-0, 3-2-1, and. 2-2-2.

6-0-0: Only one way

5-1-0: Take the ball that is alone, so 6 ways.

4-2-0: 6C2=15 ways for the two balls in their own box.

4-1-1: Again 15 ways, pick two of the six balls, but they cannot go in the box of 4.

3-3-0: 5C2=10 ways, by selecting two of the remaining five balls.

3-2-1: 6*10=60 ways by picking any of the six balls to be by itself.

2-2-2: 15 ways? First, you think they are distinguishable, yet you overcount by a factor 3!=6.

Thus, adding them all up, we get \(1+6+15+15+10+60+15=\boxed{122}.\)

See this link: https://web2.0calc.com/questions/some-questions-please-elaborate-thanks#r1

tertre Apr 16, 2019