JEE Advanced 2022 — Paper 1 — Question 7

The allowed digits are $\{0,2,3,4,6,7\}$. We count 4-digit numbers from 2022 to 4482 inclusive. For thousands digit 2: split by hundreds digit.

If hundreds digit is 0, total $6\times6=36$ numbers in 2000-2099, but we subtract the 7 numbers below 2022 (2000,2002,2003,2004,2006,2007,2020) to get $36-7=29$.

If hundreds digit is 2,3,4,6,7 (5 choices), each gives $6\times6=36$ numbers, all in interval: $5\times36=180$.

Total for thousands=2: $29+180=209$. For thousands digit 3: all $3abc$ with $a,b,c$ from the set: $6\times6\times6=216$, all in interval. For thousands digit 4: hundreds digits allowed are 0,2,3,4 (since 6,7 exceed 4482).

Each gives $6\times6=36$ numbers (44xx numbers all ≤4477≤4482): $4\times36=144$. Total: $209+216+144=569$. Thus, the number of 4-digit integers in [2022,4482] using only digits $0,2,3,4,6,7$ is 569.