JEE Main · 2023 · 12 Apr 2023 · Shift 1 · Permutations and Combinations

A number divisible by $5$ must end in $0$ or $5$, and to exceed $40000$ its leading digit must be one of $5,7,9$. When the last digit is $0$, the leading digit is chosen from $5,7,9$ in $3$ ways and the three middle places are filled from the remaining four digits in ${}^4{P}_3=24$ ways, giving $3\times 24=72$ numbers. When the last digit is $5$, the leading digit is chosen from $7,9$ in $2$ ways and the three middle places are filled from the remaining four digits in ${}^4{P}_3=24$ ways, giving $2\times 24=48$ numbers. Adding the two cases, the number of such five digit numbers is $72+48=120$.