The case of indistinguishable boxes requires Stirling's number of the 2nd kind (this arises due to a recursive nature of combinations: you put ball A into a box, then ball B can go into the box with A or one of the other two boxes, etc.):



S(n,k) = Stirling's number of the 2nd kind

S(n,k) = 



        with C(k,j) = k!/((k-j)!j!) 



We have  n=6 balls, k=3 boxes.  You can distribute the balls into one box, into two boxes, or all three boxes: 



S(6,1) + S(6,2) + S(6,3) = 1 + 31 + 90 =  







 

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