JEE Advanced · 2022 · 28 Aug 2022 · Paper 1 · Permutations and Combinations

Only the digits $\{0,2,3,4,6,7\}$ are allowed. For leading digit $2$ (numbers $2abc$) there are $6^3=216$ possibilities, of which $7$ fall below $2022$ (the six $200c$ and the single $2020$), leaving $209$. For leading digit $3$ every $3abc$ lies in range, giving $216$. For leading digit $4$: with the second digit $0,2$ or $3$ all $3\times 6^2=108$ qualify, and with second digit $4$ the largest value $4477$ is still below $4482$, so all $6^2=36$ qualify, giving $144$. The total is $209+216+144=569$.