QJEEAdv12Ma122b: JEE Advanced 2012 Mathematics - Class 11 JEEAdv12Ma122b Solution: The total number of ways in which 5 balls of different colours…

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.