Case 1:All four balls in the same box: 1 way
Since all the boxes are identical.
Case 2:Two balls in 1 box and the remaining two in another box:
=3 ways
It is divided by 2! because the boxes are identical.
Case 3:Two balls in a box and one ball in each of the remaining two boxes:
4C2=6 ways
Note: We need not select a box for the remaining two balls as they will go one each in the remaining two boxes in one way only.
Case 4:Three balls in one box and one remaining ball in another box:
4C3=4 ways
Since there is only one way of selecting the other box as the boxes are identical.
Total ways = 1 + 3 + 6 + 4 = 14