jmc

Since the balls are indistinguishable, we must only count the number of balls in the different boxes.

There are $3$ ways to arrange the balls as $(5,0,0)$ (specifically, box 1 can have 5, box 2 can have 5, box 3 can have 5).

There are $3!=6$ to arrange $(4,1,0)$ and $3!=6$ ways to arrange $(3,2,0)$; in each case, we must choose one of the 3 boxes to have the largest number of balls, and also one of the remaining two boxes to be left empty.

However, there are only $3$ ways to arrange $(3,1,1)$, and $3$ ways to arrange $(2,2,1)$; in each case, we must choose one box to have the `different' number of balls (3 in the $(3,1,1)$ case and 1 in the $(2,2,1)$ case).

This gives a total of $3+6+6+3+3=\boxed{21}$ arrangements.