Question

Phone numbers consist of a three-digit area code followed by seven digits. If the area code must have a 0 or1 for the second digit, and neither the area code nor the seven-digit number can start with 0 or 1, how many different phone numbers are possible? How did you come up with your answer?

a.
8 times 10 times 10 times 8 times 10 times 10 times 10 times 10 times 10 times 10 = 6,400,000,000
b.
8 times 2 times 10 times 8 times 10 times 10 times 10 times 10 times 10 times 10 = 1,280,000,000
c.
8 times 2 times 10 times 8 times 10 times 10 times 10 times 10 times 10 = 128,000,000
d.
8 times 2 times 10 times 8 times 10 times 10 times 10 times 10 times 10 times 10 times 10 = 12,800,000,000



Please select the best answer from the choices provided

Answer 1

The correct answer is (b) 1,280,000,000.

Step-by-step explanation:

Step-by-step explanation:

For the area code, we have two choices for the second digit (0 or 1), and 8 choices for each of the first and third digits (since they cannot be 0 or 1). Therefore, there are 8 x 2 x 8 = 128 possible area codes.

For the seven-digit number, we have 8 choices for the first digit (since it cannot be 0 or 1), and 10 choices for each of the remaining 6 digits. Therefore, there are 8 x 10 x 10 x 10 x 10 x 10 x 10 = 8 x 10^6 possible seven-digit numbers.

To get the total number of possible phone numbers, we multiply the number of possible area codes by the number of possible seven-digit numbers:

128 x 8 x 10^6 = 1,280,000,000

Therefore, there are 1,280,000,000 possible phone numbers that satisfy the given conditions.