Question

Find the total number of 9 digit numbers which have all different digits.

• A. 3265920
• B. 3267720
• C. $9\times 9!$
• D. $9\times 10!$

Answer 1

Answer: (A), (C)

$9\times 9!$

${}^{10}{P}_9-{}^9{P}_8=\frac{10!}{1!}-\frac{9!}{1}=9!(10-1)=9(9!)$
$=9\times (9\times 8\times 7\times 6!)$
$=81\times 56\times 720=3265920$.

Alternative method.
The number is to be of 9 digits. The first place can be filled in 9 ways only (as zero cannot be in the first place). Having filled up the first place the remaining 8 places can be filled up by the remaining 9 digits in ${}^9{P}_8=9!ways$.
Hence the total is $9\times 9!$.