How many numbers are there between 100 and 1000 such that at least one of their digits is 7?

Explanation

Clearly, the numbers between 100 and 1000 are 3-digit numbers.
So, we have to form 3-digit numbers such that at least one of their digits is 7.
Case I Three -digit numbers with 7 at unit's place:
The number of ways to fill the hundred's place `= 9 " " `[by any digit from 1 to 9].
The number of ways to fill the ten's place `= 10 " " `[by any digit from 0 to 9].
The number of ways to fill the unit's place `=1" " `[by 7 only].
`therefore " the number of such numbers"=(9xx10xx1) =90`.
Case II Three-digit numbers with 7 at ten's palce:
The number of ways to fill the hundred's place `= 9 " " `[by any digit from 1 to 9].
The number of ways to fill the ten's place `=1 " " `[by 7 only].
The number of ways to fill the unit place `=10 " " `[ by any digit from 0 to 9].
`therefore " the number of such numbers "=(9xx1xx10) =90.`
Case III Three-digit numbers with 7 at hundred's place:
The number of ways to fill the hundred's place `=1 " " `[by 7 only].
The number of ways of fill the ten's palce `=10 " " `[by any digit from 0 to 9].
The number of ways to fill the unit's place ` =10 " " `[by any digit from 0 to 9].
`therefore " the number of such numbers "=(1xx10xx10)=100.`
Hence, the total number of required numbers `=(90+90+100) =280.`