Question

in a different plan for area codes, the first digit could be any number from 5 through 8, the second digit was either 5, 6, or 7, and the third digit could be any number except 2. with this plan, how many different area codes are possible?
there are \boxed{} possible area codes.

Explanation:

Step1: Determine the number of choices for each digit

• For the first digit: It can be any number from 5 through 8. So there are \(8 - 5+ 1=4\) choices.
• For the second digit: It was either 5, 6, or 7. So there are 3 choices.
• For the third digit: It could be any number except 2. Since there are 10 possible digits (0 - 9) in total, there are \(10-1 = 9\) choices.

Step2: Use the multiplication principle

The multiplication principle states that if there are \(m\) ways to do one thing, \(n\) ways to do a second thing, and \(p\) ways to do a third thing, then the total number of ways to do all three things together is \(m\times n\times p\).
Here, \(m = 4\), \(n=3\), and \(p = 9\). So the total number of area - codes is \(4\times3\times9\).

$$4\times3\times9=(4\times3)\times9=12\times9 = 108$$

Answer:

108