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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
 #1
avatar+4330 
+2

 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
 #2
avatar+25 
-5

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

 Apr 24, 2019

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