JEE Main · 2021 · 25 Feb 2021 · Shift 1 · Permutations and Combinations

A number between $100$ and $1000$ formed from $1,2,3,4,5$ without repetition uses three distinct digits. Divisibility by $5$ needs $5$ in the unit place, and the other two places can be filled in $4\times 3=12$ ways. Divisibility by $3$ needs the digit sum to be a multiple of $3$, which holds for the triples $\{1,2,3\}$, $\{1,3,5\}$, $\{2,3,4\}$ and $\{3,4,5\}$, each giving $3!=6$ numbers, so $4\times 6=24$ numbers. Both conditions hold for $\{1,3,5\}$ and $\{3,4,5\}$ with $5$ in the unit place, which is $2\times 2=4$ numbers. The count is $12+24-4=32$.