Question

Find the total number of $9$ digit numbers which have all the digits different.

• A. $9\times 9!$
• B. $9!$
• C. $10!$
• D. None of these.

Answer 1

The correct option is A

$9\times 9!$

Explanation for the correct option:

Finding 9 digit number.

There are $10$different digits. Those are $0,1,2,3,4,5,6,7,8,9$.

First digit of those $9$digit numbers can't be $0$. However, the rest of the all digits can be any number.

Therefore, first digit must $9$ different numbers.

Second digit must be $9$ different numbers.

Third digit must be $8$ different numbers.

In the same way, the $9^{th}$ digit must have $2$ different numbers.

therefore, total different numbers

$$\begin{gathered} =9\times 9\times 8\times 7\times ·········\times 2\times 1 \\ =9\times 9! \end{gathered}$$

Therefore, Option(A) is the correct answer.