Explanation

(c)
Let the boxes be marked as $A,\ B,\ C$. We have to ensure that no box remains empty and all five balls have to put in. There will be two possibilities.
(i) Any two box containing one ball each and 3^{rd} box containing 3 balls. Number of ways
$= A(1)B(1)C(3)$
$= {\ ^{5}C}_{1}.^{4}C_{1}.^{3}C_{3} = 5.4.1 = 20$
Since, the box containing 3 balls could be any of the three aboxes $A,\ B,\ C$. Hence, the required number of ways $20 \times 3 = 60$
(ii) Any two box containing 2 balls each and 3rd containing 1 ball, the number of ways
$= A(2)B(2)C(1) = {\ ^{5}C}_{2}.{\ ^{3}C}_{2}.{\ ^{1}C}_{1}$
$= 10 \times 3 \times 1 = 30$
Since, the box containing 1 ball could be any of the three boxes $A,\ B,\ C$. Hence, The required number of ways
$= 30 \times 3 = 90$
Hence, total number of ways $= 60 + 90 = 150$