Question

How many area codes of the form (XYZ) are possible if the digit 'X' and 'Y' can be any number (0 through 9 but they can't repeat and the digit Z can be any number 1 through 9?

Answer 1

810 possible area codes

Step-by-step explanation

We have been told that there are 3 spots up for filling.

The first spot, X, can be any number from 0 through 9, that is, any of the 10 numbers.

The second spot, Y, can be any number from 0 through 9 too, but now, we can't repeat any number used for the first spot (X), so, we have 9 numbers that can fill this space.

For the third spot, Z, any digit from 1 through 9 can take this spot. That is any 9 numbers.

The number of area codes possible = 10 × 9 × 9 = 810 possible area codes

Hope this Helps!!!