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 ² C₁

Hence the number of total possible outcomes is ⁹C₁ × ⁹C₁ × ⁸C₁ × ⁷C₁ × ⁶C₁ = 9 × 9 × 8 × 7 × 6 × 5 × 4 × 3 × 2 = 9 (9!)