JEE Main · 2021 · 18 Mar 2021 · Shift 1 · Permutations and Combinations

The digit $3$ never occurs in $1000$, so counting over $1$ to $999$ is enough. Writing every such integer as a three-digit string with leading zeros turns the list into all $1000$ strings from $000$ to $999$, and the padding adds no further $3$s. In each of the three places the digit $3$ appears in exactly $10\times 10=100$ of these strings, because the other two places are free. The three places therefore contribute $100$ occurrences each, and the digit $3$ is written $100+100+100=300$ times.