Question

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

• A. $9!\times 2!$
• B. $10!$
• C. $9\times 9!$
• D. None of the above

Answer 1

The correct option is C $9\times 9!$

The number is to be of $9$ digits

the first place can be filled in $9$ ways only (as $0$ can not be in the left most position )

Having filled up the first place the remaining $8$ places can be filled in $9\times 8\times 7\times ...\times 1=9!$ ways

Hence total number of $9$ digit numbers with distinct digits is $=9\times 9!$