Question

Telephone numbers are preceded by a 3-digit area code (e.g. 647). In the following questions, any digit from {0, 1, 2, 3, 4, 5, 6, 7, 8, 9} can be used unless otherwise stated. How many area codes are possible if: the first digit cannot be 1 or 2 or 3? the first digit cannot be 0 or 1, the second digit must be 2 or 3, and the third digit must be 6? all three digits must be the same? e.g. 6-6-6 the first digit cannot be 0 or 1, the second digit cannot be 2 or 6, and the third digit must be 6?

Answer 1

The number of possible area codes vary depending on the rules set for each digit. We have 700, 16, 10, and 64 possible area codes for each respective rule by multiplying the number of possibilities for each digit position.

Step-by-step explanation:

To determine how many possible area codes are there based on a given set of rules, we can simply multiply the number of choices for each digit.

1. The first digit cannot be 1, 2, or 3: This means the first digit has 7 possibilities (0, 4, 5, 6, 7, 8, 9). Since there are no restrictions on the other two digits, each has 10 possibilities, giving us 7 × 10 × 10 = 700 possible area codes.
2. The first digit cannot be 0 or 1, the second digit must be 2 or 3, and the third digit must be 6: This gives us 8 possibilities for the first digit (2-9), 2 possibilities for the second (2 or 3), and only 1 possibility for the third digit (6), so 8 × 2 × 1 = 16 possible area codes.
3. All three digits must be the same: There are 10 digits possible for the first digit, and because all three must be the same, this gives us 10 possible area codes (000, 111, ..., 999).
4. The first digit cannot be 0 or 1, the second digit cannot be 2 or 6, and the third digit must be 6: The first digit has 8 possibilities (2-9), the second has 8 possibilities (0, 1, 3, 4, 5, 7, 8, 9), and the third digit is fixed, so there is only 1 possibility. This gives us 8 × 8 × 1 = 64 possible area codes.