Question

A three digit number '$n$' is such that the last two digits of it are equal and different from the first. The number of such '$n$'s is:

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

Answer 1

Answer: (C)

$81$

The hundred's place can be filled in $9$ ways. (all digits except $0$)

Once the hundred's place is filled, the ten's place can be filled in $9$ ways. (all digits except the one at hundred's place).

Finally the unit's place can be filled in only $1$ way. (since it has to be same as the digit at ten's place.)

So, using multiplication principle, as shown in figure, the total number of three digit numbers possible $=9\times 9\times 1=81$