jmc

Without regard to the distinguishability of the balls, they can be organized into groups of the following: $$(4,0,0),(3,1,0),(2,2,0),(2,1,1).$$Now we consider the distinguishability of the balls in each of these options.

(4,0,0): There is only $1$ way to do this (since the boxes are indistinguishable).

(3,1,0): There are $4$ options: we must pick the ball which goes into a box by itself.

(2,2,0): There are $\left(\genfrac{}{}{0ex}{}{4}{2}\right)=6$ ways to choose the balls for the first box, and the remaining go in the second box. However, the two pairs of balls are interchangeable, so we must divide by 2 to get $6/2=3$ arrangements.

(2,1,1): There are $\left(\genfrac{}{}{0ex}{}{4}{2}\right)=6$ options for picking the two balls to go in one box, and each of the other two balls goes into its own box.

The total number of arrangements is $1+4+3+6=\boxed{14}$.