Given : 

Nine-digit number is required in which the first digit cannot be zero and the repetition of digits is not allowed.

Assume nine boxes, 

Now, 

The first box can be filled with one of the nine available digits, so the possibility is ⁹C₁ 

Similarly,

The second box can be filled with one of the nine available digits, so the possibility is ⁹C₁ 

The third box can be filled with one of the eight available digits, so the possibility is ⁸C₁ 

The fourth box can be filled with one of the seven available digits, so the possibility is ⁷C₁ 

The fifth box can be filled with one of the six available digits, so the possibility is ⁶C₁ 

The sixth box can be filled with one of the six available digits, so the possibility is ⁵C₁ 

The seventh box can be filled with one of the six available digits, so the possibility is ⁴C₁ 

The eighth box can be filled with one of the six available digits, so the possibility is ³C₁ 

The ninth box can be filled with one of the six available digits, so the possibility is ²C₁ 

Hence,

The number of total possible outcomes is ⁹C₁ × ⁹C₁ × ⁸C₁ × ⁷C₁ × ⁶C₁ 

= 9 × 9 × 8 × 7 × 6 × 5 × 4 × 3 × 2 

= 9 (9!)