15 is the answer. We know we can split the balls into four groups: 4-0-0-0, 3-1-0-0, 2-2-0-0, 2-1-1-0, 1-1-1-1.
Case 1: 4-0-0-0 has only 1 way we can assign the balls to different boxes.
Case 2: 3-1-0-0 has (I think) 4 ways we can assign the balls to different boxes. The one ball can go anywhere, and the three other balls must go in a different box
Case 3: 2-2-0-0 has 4C2 = 6 ways to choose two balls to go in one box, then the other two balls must go in separate boxes. But since order doesn't matter, (there is no first box or second box), we divide by 2.
Case 4: 2-1-1-0 has (again) 4C2 = 6 ways to choose two balls to go in one box, then the other balls go in seperate boxes.
Case 5: 1-1-1-1 has only 1 way.
1 + 4 + 6/2 + 6 + 1 = 15.
Hope this helps! :)