1. Interpret the problem

We have:

• $5$ balls, all of different colours  so they are distinct.
• $3$ persons, also distinct.
• Each person must get at least one ball.

We need the number of onto distributions of $5$ distinct objects among $3$ distinct persons.

2. Total distributions without restriction

Each of the $5$ distinct balls can be given to any one of the $3$ persons.

So total number of distributions is $3^5 = 243.$

3. Subtract distributions where at least one person gets no ball

Let the three persons be $P_1, P_2, P_3$.

Define:

• $A_i$ = set of distributions where person $P_i$ gets no ball.

If one specific person gets no ball, then each ball can go to only the remaining $2$ persons. Thus, $|A_i| = 2^5 = 32.$

Since there are $3$ persons, $|A_1|+|A_2|+|A_3| = 3 \cdot 2^5 = 96.$

4. Add back distributions where two persons get no ball

If two specific persons get no ball, then all $5$ balls must go to the remaining one person. So for any pair, $|A_i \cap A_j| = 1.$

Number of such pairs is $\binom{3}{2}=3$. Hence, $|A_1\cap A_2|+|A_2\cap A_3|+|A_3\cap A_1| = 3.$

There is no case where all three get no ball.

5. Apply Inclusion-Exclusion Principle

Required number of distributions: $3^5 - \binom{3}{1}2^5 + \binom{3}{2}1^5$ $= 243 - 96 + 3$ $= 150.$

6. Check with options

The correct option is: $\boxed{150}$ which is Option B.

7. Comparison with stored correct answer

Stored correct answer: B

Our derived answer: B

So they agree.