Question

How many phone numbers are possible in the (770) area

code if
For the form ABC-XXXX, À is restricted to numbers 2-9. B, C,
and X can be any digit 0-9. Also, the ABC of the phone number
cannot be 911,

a. 7,000,000
b. 7.990,000
c. 7,999,999
d. 8,000,000

Answer 1

To find the number of possible phone numbers, we multiply the number of choices for each digit, considering restrictions on the first digit (A) and the sequence 911. The calculation is 8 x 10^7, then we subtract 1 for the disallowed 911 sequence, resulting in 7,999,999 possible phone numbers.

Step-by-step explanation:

The question asks how many phone numbers are possible in the (770) area code if the first digit (A) is restricted to numbers 2-9, and the other digits (B, C, X) can be any digit from 0-9, with the exception that the ABC combination cannot be 911. To calculate this, we consider each position individually.

For the first digit (A), there are 8 possible choices (2-9). For the second (B) and third (C) digits, there are 10 possible choices each (0-9), and for the four X digits, each has 10 choices as well (0-9).

The total number of combinations is therefore:

8 (A) × 10 (B) × 10 (C) × 10 (X1) × 10 (X2) × 10 (X3) × 10 (X4) = 8 × 10^7 combinations. This gives us 8,000,000 potential combinations.

However, we must subtract the combinations that contain the 911 sequence for ABC. Since there is only one combination of 911, we subtract this from the total:

8,000,000 - 1 = 7,999,999 possible phone numbers.