Let the boxes be marked as A, B, C. We have to ensure that no box remains empty and in all five balls have to put in. There will be two possibilities.

i) Any two containing one and 3rd containing 3.

A(1) B(1) C(3)

5 C 1 . 4 C 1 . 3 C 3 = 5.4.1 = 20

Since the box containing 2 each and 3rd containing 1.

A(2) B(2) C(1)

5 C 2 . 3 C 2 . 1 C 1 = 10 x 3 x 1 = 30

Since the box containing 1 ball could be any of the three boxes A, B, C

Hence the required number is = 30 x 3 = 90.

Hence total number of ways are = 60 + 90 = 150.