Question

How many phone numbers are possible in the (770) area code if the form is ABC-XXXX, where A is restricted to numbers 2-9, and B, C, and X can be any digit 0-9? Additionally, the ABC of the phone number cannot be 911.

a) 8,000,000
b) 7,990,000
c) 7,000,000
d) 7,999,999

Answer 1

To find the number of possible phone numbers, multiply the choices for each digit and subtract the combinations for restricted sequences. There are 8 choices for the first digit, 10 each for the second and third digits, and 10,000 for the last four digits. Subtract the 911 combinations to get 7,990,000 possible phone numbers.

Step-by-step explanation:

The question asks us to calculate the number of possible phone numbers in the (770) area code, given the constraints on the digits that can be used. For the first digit, 'A', we have 8 choices (2-9). For the second and third digits, 'B' and 'C', we have 10 choices each (0-9). Each of the four 'X' digits also has 10 choices (0-9). However, we must exclude the sequence 911. Therefore, we calculate the total number like this:

• Possible combinations for 'A': 8 choices
• Possible combinations for 'B' and 'C': 10 choices each
• Possible combinations for 'XXXX': 10 choices each, raised to the power of 4

So, we have 8 * 10 * 10 * (10^4) = 8,000,000 possible phone numbers. But we must subtract the combinations where 'ABC' is 911, which amount to 1 * 1 * 1 * (10^4) = 10,000 combinations. Thus, the correct number is 8,000,000 - 10,000 = 7,990,000 possible phone numbers.