In forming the nine-digit numbers, the first digit, excluding zero, can be one of the 9 and the remaining 8 digits, together with zero, making up 9 digits may be used in forming the next 8 digits in 9P8 = 9! ways.
Hence, the required number of 9-digit numbers = 9(9!)
Now regarding their sum:
With 1 in the units digit there are 8 x 8P7 = 8(8!) numbers; and the 1 in all these numbers added up make up a sum 8(8!). The same is true of numbers with 2 in the units digit, but their sum is 8(8!) 2. This way the sum of all the numbers in the unit digit is 8(8!) (1 + 2 + … + 9) = 8(8!) 45. If any digit, instead of being in the unit place, is in the tenths place the value will be 10.
Proceeding in the same way, sum of all nine-digit number with different digits is
8 x 8! x 45(1 + 10 + 102 + … + 108)
