JEE Main 2025 Apr 2 Shift 2, Mathematics Q16: Combinations

Let the rows contain $n_1=3,n_2=3,n_3=2$ boxes respectively (total $8$). Place the $5$ distinct letters into the $8$ boxes with at most one letter per box.

Total placements (no row restriction): choose $5$ of the $8$ boxes and arrange the letters in $5!$ ways $\Rightarrow {}^8{C}_5\cdot 5!=56\cdot 120$.

Subtract the placements that leave at least one row empty using inclusion-exclusion. The number of placements with row $i$ empty (all 5 letters land in the other two rows, which together have $8-n_i$ boxes):

Row 1 empty: ${}^5{C}_5\cdot 5!=5!$.

Row 2 empty: ${}^5{C}_5\cdot 5!=5!$.

Row 3 empty: ${}^6{C}_5\cdot 5!=6\cdot 5!$.

Two rows simultaneously empty would force all $5$ letters into a single row with at most $3$ boxes — impossible, so intersection terms vanish.

Valid placements $=56\cdot 5!-(5!+6\cdot 5!+5!)=5!(56-1-6-1)=120\cdot 48=5760$.