In a different plan for area codes, the first digit could be any number from 3 through 6 , the second digit was either 7 or 8 , and the third digit could be any number except 2 or 6 . With this plan, how many different area codes are possible? There are \( \square \) possible area codes.

To find the total number of different area codes possible under the described plan, we'll analyze the options for each digit: 1. For the **first digit**, it can be any number from 3 to 6. That gives us **4 options**: {3, 4, 5, 6}. 2. For the **second digit**, it can only be 7 or 8. That gives us **2 options**: {7, 8}. 3. For the **third digit**, it can be any digit from 0 to 9 except for 2 and 6. This means there are 8 valid digits, specifically {0, 1, 3, 4, 5, 7, 8, 9}. Now we multiply the number of options for each position: \[ 4 \text{ (first digit)} \times 2 \text{ (second digit)} \times 8 \text{ (third digit)} = 4 \times 2 \times 8 = 64. \] Thus, there are **64 possible area codes** under this plan.