Question

Current rules for telephone area codes allow the use of digits​ 2-9 for the first​ digit, and​ 0-9 for the second and third​ digits, but the last two digits cannot both be 1​ (to avoid confusion with area codes such as​ 911). How many different area codes are possible with these​ rules? That same rule applies to the exchange​ numbers, which are the three digits immediately preceding the last four digits of a phone number. Given both of those​ rules, how many​ 10-digit phone numbers are​ possible? Given that these rules apply to the United States and Canada and a few​ islands, are there enough possible phone​ numbers? (Assume that the combined population is about​ 400,000,000.)

Answer 1

Using the fundamental counting theorem, we have that:

• 648 different area codes are possible with this rule.
• There are 6,480,000,000 possible 10-digit phone numbers.
• The amount of possible phone numbers is greater than 400,000,000, thus, there are enough possible phone numbers.

The fundamental counting principle states that if there are p ways to do a thing, and q ways to do another thing, and these two things are independent, there are ways to do both things.

For the area code:

• 8 options for the first digit.
• 9 options for the second and third.

Thus:

$8 \times 9 \times 9 = 648$

648 different area codes are possible with this rule.

For the number of 10-digit phone numbers:

• 7 digits, each with 10 options.
• 648 different area codes.

Then

$648 \times 10^7 = 6,480,000,000$

There are 6,480,000,000 possible 10-digit phone numbers.

The amount of possible phone numbers is greater than 400,000,000, thus, there are enough possible phone numbers.

A similar problem is given at brainly.com/question/24067651