Question

How many three-digit numbers are there with $3$ in the unit place? (i) with repetition (ii) without repetition.

Answer 1

$⟹$ Unit digit is $3$

$\left(i\right)$ With repetition

Hundreds Digits has $9$ possible chances excluding $0$

Tens Digit has $10$

So possible three-digit numbers is $9\times 10=90$

$\left(ii\right)$ without repetition

Hundreds Digits has $9$ possible chances excluding $0$

Tens Digit has $9$ excluding the one that is in Hundered`s place

So possible three-digit numbers is $9\times 9=81$