Question

Find the number of 9-digit codes possible if the second digit cannot be a four.

Answer 1

As per given by the question,

There are given that 9- digit codes.

So,

Here, take 0 to 8.

So, total number is,

$9^9$

Now,

From the question, there are clearly mention that second digit cannot be 4.

If second digit cannot be 4, then this is not a 9 digit number.

So,

The number is,

$9^8$

Then,

The total number is,

$\begin{gathered} 9^9-9^8=9^8(9-1) \\ =8\times9^8 \end{gathered}$

Hence, the total number is,

$8\times9^8$