JEE Advanced 2022 · Paper 1, Mathematics question 7

The digits $0,2,3,4,6,7$ may repeat.

First digit 2 (2000-2999): $6^3=216$ numbers. Those below 2022 are 2000, 2002, 2003, 2004, 2006, 2007 and 2020, so we keep $216-7=209$.

First digit 3: all $6^3=216$ numbers.

First digit 4 (at most 4482):

• second digit 0, 2 or 3: $3\times36=108$;
• second digit 4: every third digit is at most 7, so all of $44xy$ are at most 4482: 36.

Total. $209+216+108+36=569$.