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

 

2. How many ways are there to put 5 balls in 3 boxes if the balls are not distinguishable and neither are the boxes?

 Feb 15, 2019
 #1
avatar+101798 
+3

Assuming that some boxes could remain empty.....

 

For the first one....let the number of balls = k   and the number of boxes = n

 

So...we have

 

C( k + n - 1 , n - 1)   = C( 6 + 3 - 1 , 3 - 1)  = C (8, 2)  = 28 ways

 

 

 

For the second one.....we have the following partitions of 5 indistinguishable balls in 3 indistinguishable  boxes

 

5, 0, 0

4, 1, 0

3, 2, 0

3, 1, 1

2, 2, 1   =  5 ways

 

 

cool cool cool

 Feb 15, 2019
edited by CPhill  Feb 15, 2019
 #2
avatar+4296 
+3

Yes, this method is called "Stars and Bars," or "Balls and Urns."

tertre  Feb 16, 2019

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