Question -
Answer -
In a nine-digitnumber, 0 cannot appear in the first digit and the repetition of digits is notallowed. So, the number of ways of filling up the first digit is 9C1=9
Now, 9 digits are leftincluding 0. So, the second digit can be filled with any of the remaining 9digits in 9 ways.
Similarly, thethird box can be filled with one of the eight available digits, so thepossibility is 8C1
The fourth digitcan be filled with one of the seven available digits, so the possibilityis 7C1
The fifth digitcan be filled with one of the six available digits, so the possibility is 6C1
The sixth digitcan be filled with one of the six available digits, so the possibility is 5C1
The seventh digitcan be filled with one of the six available digits, so the possibility is 4C1
The eighth digitcan be filled with one of the six available digits, so the possibility is 3C1
The ninth digitcan be filled with one of the six available digits, so the possibility is 2 C1
Hence the number oftotal possible outcomes is 9C1 × 9C1 × 8C1 × 7C1 × 6C1 =9 × 9 × 8 × 7 × 6 × 5 × 4 × 3 × 2 = 9 (9!)