Question

How many possible $9$-digit codes can be formed?

Answer 1

Solve for $9$ digit codes

As the number is to be of 9 digits,

So we can form any number by $0,1,2,3,4,5,6,7,8$and $9$digits.

In order to form a $9$ digit number, the first digit should be free from $0$.

Therefore the first place, we have $9$ choices only.

So the first place can be filled in $9$ ways.

After filling up the first place, each remaining place can be filled by $10$ digits.

$\therefore$The number of ways is $9\times 10\times 10\times 10\times 10\times 10\times 10\times 10\times 10\times 10=9\times {10}^8$

Hence the possible number of $9$digit codes is $9\times {10}^8$