Question

The total number of 9-digits 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-digits numbers, having all digit are different $=9{}^9{P}_8$
$=\frac{9\times 9!}{1!}=9\times 9!$
$(\because$ In first place, we select anyone of nine numbers except zero. In rest of the eight places, we select any eight numbers from remaining 9 numbers)