How many ways are there to put 4 balls in 3 boxes if the balls are not distinguishable but the boxes are?
Since the balls are indistinguishable, we only have to consider the number of balls in the boxes.  The arrangements for balls in boxes are $$(4,0,0),(3,1,0),(2,2,0),(2,1,1).$$However, since the boxes are distinguishable, we must also consider the arrangement of balls in the boxes in order.

For (4,0,0), there are $3$ different ways (box $\#1$ can have 4, box $\#2$ can have 4, or box $\#3$ can have 4).

For (3,1,0), there are $3! = 6$ ways: we have 3 choices for the box containing 3 balls, then 2 choices for the box containing 1 ball.

For (2,2,0) there are $3$ ways: we must choose the box which remains empty.

For (2,1,1) there are $3$ ways: we must choose the box which gets 2 balls.

This gives a total of $3 + 6 + 3 + 3 = \boxed{15}$ arrangements.