How many 3 digit numbers can be made if no digit is repeated and the number can not begin with a zero?

digit #1 - 9 numbers (1-9)

digit #2 = 9 numbers (0-9 minus the first digit)

digit #3 = 8 numbers (0-9 minus the first 2 digits)

9*9*8 = 648 combinations

We have to select 3 digits out of 10 with the following conditions:

1) It can't start with 0

2) repetition is not allowed.

Let ABC this number:

A = 9 choices (out of 10, excluding 0, because it can't be 0)

B = 9 choices (out of 10 including 0, because it can be 0)

C = 8 choices (out of 10 because 2 digits already selected)

The total ways to write this number are:

9 x 9 x 8 = 648 ways

Another method:

A can be written in ⁹C₁

B can be written in ⁹C₁

C can be written in ⁸C₁

Total ways: ⁹C₁ x ⁹C₁ x ⁸C₁ = 648 ways