Here, 5 distinct balls are to be distributed amongst 3 persons so that each gets at least one ball. So, two possible cases arises

Case I : Two of the persons get one-one ball each and the third person gets three balls.

i.e.

Now, A can get the ball in $${ }^5 \mathrm{C}_1$$ ways. After that, $$B$$ can get one ball in $${ }^4 C_1$$ ways and then after $$C$$ can get three balls in $${ }^3 \mathrm{C}_3$$ ways.

Hence, Total number of ways

$$\begin{aligned} & ={ }^5 \mathrm{C}_1 \cdot{ }^4 \mathrm{C}_1 \cdot{ }^3 \mathrm{C}_3 \times \frac{3!}{2!} \\ & =5 \times 4 \times 1 \times \frac{3 \times 2 \times 1}{2 \times 1} \\ & =60 \end{aligned}$$

Case II : Two of the persons get two-two balls each and the third person gets one ball.

i.e.

Now, A can get two balls in $${ }^5 \mathrm{C}_2$$ ways. After that, $$B$$ can get 2 ball in $${ }^3 \mathrm{C}_2$$ ways and then after $$C$$ can get 1 ball in $${ }^1 C_1$$ way. Hence, total number of ways

$$\begin{aligned} & ={ }^5 C_2 \cdot{ }^3 C_2{ }^1 C_1 \cdot \frac{3!}{2!} \\ & =\frac{5 \times 4}{2 \times 1} \times \frac{3 \times 2}{2 \times 1} \times 1 \times \frac{3 \times 2 \times 1}{2 \times 1} \\ & =90 \end{aligned}$$

Hence, total number of ways to distribute 5 balls $$=60+90=150$$

Combination with Repetition

(i) Only two possible cases arises:

Case I :

Case II:

(iii) Use the concept of combination to find individual ways and then add up the total ways in each case.