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?

Answer 1

Let's start with the three digit area code first.
We have three slots to fill.

The first slot cannot start with 0 or 1, which leaves us 2, 3, 4 ..., 9 to fill.
This means we have 8 different arrangements for the first slot.

The second slot MUST have a 0 or 1 for the second digit, which leaves us with 2 different arrangements.

There are no restrictions for the third slot, so we would have 10 different arrangements.

Thus, the total arrangement of area code is 8 · 2 · 10 = 160 different arrangements.

Now, let's explore the total arrangements of telephone numbers.
There is only one restriction, and that's on the first number.

Thus, the total number of ways is:
8 · 10 · 10 · 10 · 10 · 10 · 10 = 8, 000, 000

Since we want them simultaneously, we want to multiply the two arrangements to find the total number of arrangements.

Total arrangements: 160 · 8, 000, 000 = 1 280 000 000 different ways.

Answer 2

In this exercise we have to use the knowledge of probability to find the phone number, so:

1 280 000 000 different ways.

First we have to establish some information about this exercise such as:

• The first slot cannot start with 0 or 1, which leaves us 2, 3, 4 ..., 9 to fill.

• This means we have 8 different arrangements for the first slot.

• The second slot have a 0 or 1 for the second digit, which leaves us with 2 different arrangements.

• There are no restrictions for the third slot, so we would have 10 different arrangements.

Thus, now calculating the possibilities of occurrence, we have:

$8 * 2 * 10 = 160\\8* 10^5 = 8, 000, 000\\160 * 8, 000, 000 = 1 280 000 000$

See more about probability at brainly.com/question/795909