Solution :
$$\because $$ Each person gets at least one ball.
∴ 3 Persons can have 5 balls in the following systems
| Person |
I |
II |
III |
| No. of balls |
1 |
1 |
3 |
or
| Person |
I |
II |
III |
| No. of balls |
1 |
2 |
2 |
The number of ways to distribute the balls in first system
$$ = {\,^5}{C_1} \times {\,^4}{C_1} \times {\,^3}{C_3}$$
Also 3, persons having 1, 1 and 3 balls can be arranged $$\frac{{3!}}{{2!}}\,\,{\text{ways}}{\text{.}}$$
∴ No. of ways to distribute 1, 1, 3 balls to the three persons
$$ = {\,^5}{C_1} \times {\,^4}{C_1} \times {\,^3}{C_3} \times \frac{{3!}}{{2!}} = 60$$
Similarly the total no. of ways to distribute 1, 2, 2 balls to the three persons
$$ = {\,^5}{C_1} \times {\,^4}{C_2} \times {\,^2}{C_2} \times \frac{{3!}}{{2!}} = 90$$
∴ The required number of ways = 60 + 90 = 150