JEE Advanced · Mathematics · 4. P&C

The total number of ways in which 5 balls of different colours can be distributed among 3 persons so that each person gets at least one ball is

A 75
B 150
C 210
D 243

Verified Solution

Answer & Solution

Correct Answer

(B) 150

Step-by-step Solution

Detailed explanation

$\because$ Each person gets at least one ball.
$\therefore 3$ 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.
$\therefore$ 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 $={ }^{5} \mathrm{C}_{1} \times{ }^{4} \mathrm{C}_{2} \times{ }^{2} \mathrm{C}_{2} \times \frac{3 !}{2 !}=90$
$\therefore$ The required number of ways $=60+90=150$

Apply the relevant identity from the chapter and substitute the values established in the previous step, keeping the original constraint in view.

Use the simplified expression to isolate the unknown, then plug the result back into the question setup to confirm the working is internally consistent.

Cross-check against a boundary or limiting case. Both the dimensional balance and the qualitative behaviour at the extreme should agree with the candidate answer.

Combine the steps above to identify the correct option. The remaining choices each fail at least one of the consistency, dimensional, or boundary checks.

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