Question

The total number of $9$ - digit numbers which have all different digits is

• A. $10!$
• B. $9!$
• C. $9\times 9!$
• D. $10\times 10!$

Answer 1

Answer: (C)

The total number of $9$- digit numbers, having all digits are different
$=9\times {}^9{P}_8$
$=\frac{9\times 9!}{1!}$
$=9\times 9!$