In a plan for area codes, the first digit could be any number from 3 though 9, the second digit was either 0 or 4, and the third digit could be any number except 0. with this plan, how many different area codes were possible?

Step-by-step explanation:

Given that the first digit could be any number from though .

So,there are ways to choose the first digit.

Given that the second digit was either or .

So,there are ways to choose the second digit.

Given that the third digit could be any number except .

So,there are ways to choose the third digit.

The number of different codes possible is the product of number of ways of choosing first digit,number of ways of choosing second digit and number of ways of choosing third digit.

Total ways are ways

it's equivalent, so we can say that its equal.

let's call the denominator x and the numerador is 18 less then denominator, x - 18.

multiply

2.x = 5.(x - 18)

2x = 5x - 90

90 = 5x - 2x

90 = 3x

x = 90/3

x = 30

so: