The total number of ways in which 5 balls of different colours can be distributed among 3 persons so that each person gets at least one ball is

$5 \text{ distinct balls},\qquad 3 \text{ persons}$ $\text{each person gets at least one ball}$ $\text{To find: total number of such distributions}$ $(1)\ 75 \qquad (2)\ 150 \qquad (3)\ 210 \qquad (4)\ 243$ $1.\ \text{total distributions (no restriction)}$ $2.\ \text{subtract cases with someone empty}$ $3.\ \text{inclusion--exclusion}$ $3^5 = 243$ $|A_i| = 2^5 = 32,\qquad 3\,|A_i| = 3 \times 32 = 96$ $|A_i \cap A_j| = 1^5 = 1,\qquad \binom{3}{2} = 3$ $|A_1 \cup A_2 \cup A_3| = 3 \times 32 - 3 \times 1 = 96 - 3 = 93$ $3^5 - |A_1 \cup A_2 \cup A_3| = 243 - 93 = 150$ $(3,1,1):\ \binom{5}{3} = 10, \qquad (2,2,1):\ 5 \times 3 = 15$ $25 \times 3! = 25 \times 6 = 150\ \ \checkmark$ $\therefore\ \text{total} = 150 \Rightarrow \boxed{\text{option (2)}}$