jmc

In this problem we don't care which box is which, we only care which balls are together and which ones aren't.

For each ball, there are 2 choices of which box to place it in. Since this choice is independent for each of the 4 balls, we multiply the number of choices together. Hence there are $2^4=16$ ways to place 4 distinguishable balls into 2 distinguishable boxes.

We then divide by the number of ways to arrange the boxes. There are $2!=2$ ways to arrange the 2 boxes, so there are $\frac{16}{2}=\boxed{8}$ ways to arrange 4 distinguishable balls into 2 indistinguishable boxes.

Note: This method does not generalize if there are more than 2 boxes.