JEE Main 2025 Jan 23 Shift 2, Mathematics Q21: Permutation

Let $A$ be the event "all 5 boys sit together" and $B$ the event "no two boys sit together". The two events are mutually exclusive (if all 5 boys are in one block then certainly some pair of boys are adjacent).

Count $A$: treat the 5 boys as one block. There are then $5$ items (the block plus 4 girls) in a row: $5!$ arrangements. Inside the block, the boys can be ordered in $5!$ ways. So $|A|=5!\times 5!=120\times 120=14400$.

Count $B$: first seat the 4 girls in a row: $4!$ ways. This creates $5$ gaps (including ends) into which the 5 boys are placed, one boy per gap so no two boys are adjacent: $5!$ ways. So $|B|=4!\times 5!=24\times 120=2880$.

Total $=14400+2880=17280$.