How many ways are there to put 4 indistinguishable balls in 3 distinsuishable boxes?
Looks like three ways to me.
1 1 2
1 2 1
2 1 1
I do not know the answer but I think there are other ways like
4 0 0
0 4 0
3 0 1
1 0 3
2 2 0
0 2 2 etc
With no restrictions and empty boxes allowed, we have:
[4 + 3 - 1] nCr [3 - 1] = 6 C 2 = 15 ways
+37159 
