Question

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

• A. $9!$
• B. $8!$
• C. $9\times 9!$
• D. None of these

Answer 1

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

total number of digits =$10$

i.e. $0,1,2,3,4,5,6,7,8,9$

require $9$ diffrent number

$0$ cant be placed in first place

= first place can be filled in $9$ ways

and the rest $9$ blank with $9$ digits in $9!$ ways

total ways = $9\times 9!$

option C