$\because \quad$ Each person gets at least one ball.

$\therefore \quad 3$ Persons can have 5 balls as follow.

\begin{array}{|c|c|c|}

\hline Person & No. of balls & No. of balls \\

\hline I & 1 & 1 \\

\hline II & 1 & 2 \\

\hline III & 3 & 2 \\

\hline

\end{array}

The number of ways to distribute balls 1,1,3 in first to three persons

$={ }^{5} \mathrm{C}_{1} \times{ }^{4} \mathrm{C}_{1} \times{ }^{3} \mathrm{C}_{3}$

Also 3 , persons having 1,1 and 3 balls can be arranged in $\frac{3 !}{2 !}$ ways.

$\therefore \quad$ Total no. of ways to distribute $1,1,3$ balls to the three persons

$={ }^{5} \mathrm{C}_{1} \times{ }^{4} \mathrm{C}_{1} \times{ }^{3} \mathrm{C}_{3} \times \frac{3 !}{2 !}=60$

Similarly, total no. of ways to distribute $1,2,2$ balls to

three persons $={ }^{5} \mathrm{C}_{1} \times{ }^{4} \mathrm{C}_{2} \times{ }^{2} \mathrm{C}_{2} \times \frac{3 !}{2 !}=90$

$\therefore \quad$ The required number of ways $=60+90=150$