Question

How many 9 - digit numbers of different digits can be formed?

Answer 1

In a nine-digit number 0 cannot appear in the first-digit. So, the number of ways of filling up the first digit = 9.
Now, 9 digit are left including 0. So, second digit can be filled with any of the remaining 9 digits in 9 ways. Similarly, remaining digits can be filled in 8, 7,6, 5, 4,3 and 2 ways.
Hence, the total number of required numbers
= $9\times 9\times 8\times 7\times 6\times 5\times 4\times 3\times 2$
= $9\times (9!)$