Question

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

• A. $10(9!)$
• B. $8(9!)$
• C. $9(9!)$
• D. none of these

Answer 1

The correct option is C $9(9!)$

All number are formed by using the digits $0,1,2,3,4,5,6,7,8,9$
The number of arrangements each consisting of $9$ digits ${=}^{10}{P}_9=10!$
These arrangements includes those which start from $0$,

clearly all such numbers will be $8$ digit number.
the number of such arrangement ${=}^9{P}_8=9!$
Therefor number of all $9$ digit number $=10!-9!=10(9!)-9!=9\times 9!$