JEE Mathematics 2025 Question with Solution

Given: There are $5$ boys and $4$ girls.

Find: The number of seating arrangements in a row such that either all the boys sit together or no two boys sit together.

We use two separate cases and add the results.

Case 1: All boys sit together

Treat the $5$ boys as one block. Then we have a total of $5$ units: one boy-block and $4$ girls.

The number of ways to arrange these $5$ units is

$$5! = 120$$

Within the boy-block, the $5$ boys can be arranged in

$$5! = 120$$

Therefore, the total number of arrangements in this case is

$$5! \times 5! = 120 \times 120 = 14400$$

Case 2: No two boys sit together

First arrange the $4$ girls.

$$4! = 24$$

This creates $5$ gaps: before the first girl, between consecutive girls, and after the last girl.

Since no two boys can sit together and there are $5$ boys, exactly one boy must be placed in each gap.

The $5$ boys can be arranged in these $5$ gaps in

$$5! = 120$$

ways.

Therefore, the total number of arrangements in this case is

$$4! \times 5! = 24 \times 120 = 2880$$

Adding both cases,

$$14400 + 2880 = 17280$$

Therefore, the required number of ways is $17280$.

Given: $5$ boys and $4$ girls are to sit in a row.

Find: The total number of arrangements satisfying the stated condition.

The condition splits naturally into two non-overlapping cases.

1. If all boys are together, make one block of boys. Then arrange that block with the $4$ girls, and multiply by the internal arrangements of the boys.
2. If no two boys are together, arrange the girls first. The $4$ girls create exactly $5$ gaps, and with $5$ boys, each gap must receive exactly one boy.

So directly,

$$(5!)(5!) + (4!)(5!)$$

Now substitute values:

$$(120)(120) + (24)(120) = 14400 + 2880 = 17280$$

Therefore, the required answer is $17280$.