The number of times the digit 3 will be written when listing the integers from 1 to 1000 is

Since 3 does not occur in 1000. So, we have to count the number of times 3 occurs when we list the integers from 1 to 999. Any number between 1 and 999 is of the form abc, where

$0\le a,b,c\le 9$.

Clearly, Number of times 3 occurs 

= (No. of numbers in which 3 occurs exactly at one place)

 + 2 (No. of numbers in which 3 occurs exactly at two places) 

+ 3 (No. of numbers in which 3 occurs exactly at three places)

 Now, 

,No. of numbers in which 3 occurs exactly at one place: 

Since 3 can occur at one place in ${ }^3{C}_1$ ways and each of the remaining two places can be filled in 9 ways. So, number of numbers in which 3 occurs exactly at one place ${=}^3{C}_1\times 9\times 9$ .No. of numbers in which .3 occurs exactly at two places: 

Since 3 can occur exactly at two places in ${ }^3{C}_2$ ways and the remaining place can be filled in 9 ways. 

So, number of numbers in which 3 occurs exactly at two places 

No. of  numbers in which 3 occurs at all the three places ${=}^3{C}_2\times 9$

Since 3 can occur in all the three digits in one way only. So, number of numbers in which 3 occurs all the three places is one.
Hence, Numbers of times 3 occurs

${=}^3{C}_1\times 9\times 9+2\left({ }^3{C}_2\times 9\right)+3\times 1=300$