Question

How many numbers are there between $99$ and $1000$ which have exactly one of their digits as $7$?

Answer 1

Let a 3-digit number $ABC$

Case (1)

If $A=7$, then $B$ and $C$ can be any digit but not $7$

so, total $9$ digits $(0,1,2,3,4,5,6,8,9)$ can replace $B$ and $C$

Then total such combinations $=1\left(9\right)\left(9\right)=81$

Case (2)

If $B=7$ then $A$ can be any digit but not $0$ and $7$, because if $A=0$ it will be a 2-digit number.

And $C$ can be any digit but not $7$

Then total combinations $=\left(8\right)\left(1\right)\left(9\right)=72$

Case (3)

If $C=7$ then $A$ can be any digit but not $0$ and $7$, because if $A=0$ it will be a 2-digit number.

And $B$ can be any digit but not $7$

Then total combinations $=\left(8\right)\left(9\right)\left(1\right)=72$

So, total numbers between $99$ and $1000$ which have exactly one of their digits as $7$ are $=81+72+72=225$