Find the total number of 9 digit numbers which have all different digits.
3265920
9×9!
10P9−9P8=10!1!−9!1=9!(10−1)=9(9!)
=9×(9×8×7×6!)
=81×56×720=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 9P8=9! ways.
Hence the total is 9×9!.