easy

$\color{#D61F06}{\text{Let the 3 digit number be }} \overline{ABC}$

Then, there are $9$ choices for digit $A$ because it can be anything out of $(1,2,3,...,9)$ ( 0 is not allowed as A because then number is not 3 digit)

Then, there are again $9$ choices for digit $B$ , because you can't take the digit which was chosen for $A$ but you can take $0$ this time, giving $9$ choices,

Now, for $C$ , you have $8$ choices because 2 out of the total $10$ digits have already been chosen.

Thus by the $\color{#3D99F6}{Multiplication}$ $\color{#20A900}{Rule}$ , there will be $9\times 9\times 8= \boxed{\color{#69047E}{648}}$ numbers.