JEE Main · 2019 · 08 Apr 2019 · Shift 2 · Permutations and Combinations

Count the four-digit numbers greater than $4321$ formed from $\{0,1,2,3,4,5\}$ with repetition allowed. If the first digit is $5$, every choice of the other three works, giving $6^3=216$ numbers. If the first digit is $4$, look at the second digit: starting $45$ or $44$ allows anything in the last two places, $36+36=72$ numbers; starting $43$ we need the last two digits to exceed $21$, which gives $18$ numbers when the third digit is $3,4$ or $5$, plus $4$ more from $432$ with the last digit $2,3,4$ or $5$. First digits of $1,2$ or $3$ are all too small. Adding up, $216+36+36+18+4=310$.