Solution:
In a nine-digit number, there can’t be a 0 in the initial digit, and digits can’t be repeated. As a result, there are 9C1= 9 different methods to fill in the initial digit.
There are now 9 digits left, including 0. As a result, the second digit can be filled in nine different ways using any of the remaining nine digits.
Similarly, the third box can be filled with any of the eight digits available, therefore 8C1 is an option.
The fourth digit can be filled with any of the seven digits available, therefore 7C1 is a possibility.
The fifth digit can be filled with any of the six digits available, therefore 6C1 is a possibility.
The sixth digit can be filled with any of the six digits available, therefore 5C1 is a possibility.
The seventh digit can be supplied with any of the six digits available, therefore 4C1 is a possibility.
The eighth digit can be supplied with any of the six digits available, therefore 3C1is a possibility.
The ninth digit can be filled with any of the six digits available, therefore 2 C1 is a possibility.
As a result, the total number of possible outcomes is 9C1 × 9C1 × 8C1 × 7C1 × 6C1 = 9 × 9 × 8 × 7 × 6 × 5 × 4 × 3 × 2 = 9 (9!)