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Easy problem (hopefully).

 

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

 

Thanks in advance!

 Feb 10, 2019
 #1
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Assuming that we can have empty boxes

 

Let k  = the number of balls

Let n = the number of boxes

 

The total ways  =

 

C ( k + n - 1 , n - 1 )    =

 

C ( 6 + 3 - 1 , 3 - 1 )  =

 

C ( 8 , 2 )  =

 

28 ways

 

 

 

cool cool cool

 Feb 10, 2019

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