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 =