Question

A three-digit number $n$ is such that the last two digits of it are equal and differ from the first. The number of such $n\text{'}s$ is

• A. $64$
• B. $72$
• C. $81$
• D. $900$

Answer 1

The correct option is C

$81$

Find the number of ways to write required three digit number $n$:

In a three digit number, the hundredth place can be filled with digits $1$ to $9$ only, since $0$ is excluded as otherwise it would make it a two digit number.

Now, the digit at tens place can be filled in $9$ different ways (including $0$) and excluding the digit that was included in hundreds place as the last two digits differ from the first digit.

Now the unit place can be filled in only one way since the last two digits are equal and differ from the first digit.

$\therefore$total number of ways to write such a number $=9\times 9\times 1$

$=81$

Hence, option C is correct.