Solution:

Given, the number is of 9 digits and all digits are different.
Let us solve this problem step by step.

Step 1: Find the total number of digits possible for the 1st place
There are 9 digits available for the 1st place (since 0 cannot be used as the first digit).

Step 2: Find the total number of digits possible for the 2nd place
Since 1 digit is already used in the 1st place, there are only 9 digits left for the 2nd place.

Step 3: Find the total number of digits possible for the 3rd place
Since 2 digits are already used in the 1st and 2nd places, there are only 8 digits left for the 3rd place.

Step 4: Find the total number of digits possible for the 4th place
Since 3 digits are already used in the 1st, 2nd, and 3rd places, there are only 7 digits left for the 4th place.

Step 5: Find the total number of digits possible for the 5th place
Since 4 digits are already used in the 1st, 2nd, 3rd, and 4th places, there are only 6 digits left for the 5th place.

Step 6: Find the total number of digits possible for the 6th place
Since 5 digits are already used in the 1st, 2nd, 3rd, 4th, and 5th places, there are only 5 digits left for the 6th place.

Step 7: Find the total number of digits possible for the 7th place
Since 6 digits are already used in the 1st, 2nd, 3rd, 4th, 5th, and 6th places, there are only 4 digits left for the 7th place.

Step 8: Find the total number of digits possible for the 8th place
Since 7 digits are already used in the 1st, 2nd, 3rd, 4th, 5th, 6th, and 7th places, there are only 3 digits left for the 8th place.

Step 9: Find the total number of digits possible for the 9th place
Since 8 digits are already used in the 1st, 2nd, 3rd, 4th, 5th, 6th, 7th, and 8th places, there are only 2 digits left for the 9th place.

Step 10: Find the total number of 9-digit numbers with all different digits
By the multiplication principle, the total number of 9-digit numbers with all different digits is:

9 × 9 × 8 × 7 × 6 × 5 × 4 × 3 × 2 = 9! = 362880

Therefore, the correct option is A) 9 × 9!