We use cookies to personalise content and advertisements and to analyse access to our website. Furthermore, our partners for online advertising receive pseudonymised information about your use of our website. cookie policy and privacy policy.

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

 Apr 16, 2019

 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

 Apr 16, 2019

Thank you so much!!!!!!!!

 Apr 24, 2019

33 Online Users