Permutations and Combinations – Arrangements – JEE Main 02 April 2025 Shift 2

Solution:

Let the three rows of boxes be denoted as $R_1$, $R_2$, and $R_3$.
Based on the solution combinations, the rows contain $3$, $2$, and $3$ boxes respectively, making a total of $8$ boxes.

Total ways to place $5$ distinct letters in $8$ boxes (with at most one letter per box):
$$\text{Total Ways} = {}^{8}C_{5} \times 5!$$

We need to subtract the unwanted cases where at least one row remains empty.

Case 1: Row $R_3$ is empty.
All $5$ letters must be placed in rows $R_1$ and $R_2$.
Total boxes in $R_1$ and $R_2$ = $3 + 2 = 5$.
Ways to place = ${}^{5}C_{5} \times 5! = 5!$

Case 2: Row $R_1$ is empty.
All $5$ letters must be placed in rows $R_2$ and $R_3$.
Total boxes in $R_2$ and $R_3$ = $2 + 3 = 5$.
Ways to place = ${}^{5}C_{5} \times 5! = 5!$

Case 3: Row $R_2$ is empty.
All $5$ letters must be placed in rows $R_1$ and $R_3$.
Total boxes in $R_1$ and $R_3$ = $3 + 3 = 6$.
Ways to place = ${}^{6}C_{5} \times 5!$

Number of required valid ways = Total Ways – (Sum of ways with any row empty)
$$\text{Valid Ways} = {}^{8}C_{5} \times 5! – \left( 5! + 5! + {}^{6}C_{5} \times 5! \right)$$

$$\text{Valid Ways} = 56 \times 120 – \left( 120 + 120 + 6 \times 120 \right)$$
$$\text{Valid Ways} = 120 \times [56 – (1 + 1 + 6)]$$
$$\text{Valid Ways} = 120 \times (56 – 8)$$
$$\text{Valid Ways} = 120 \times 48 = 5760$$

Ans. (4)