How many phone numbers can be made if the first digit must be 1, the second digit mustbe a number in the range 3-5, the third digit must be a number in the range (6-9), and the last seven digits can be any single digit number 0-9?

136, 137, 138, 139. I observed that for every second digit there exists 4 digits that can be next, which means that there are 12 ways of arranging the first 3 digits.

Then I used the Combinations Formula to determine, in how many ways can you rewrite a string of length 7 with 10 distinct entries.

[tex]C(n, k) = \frac{n!}{k!(n-k)!}[/tex], where the '!' symbolizes the factorial operator, which returns the amount of ways to arrange n objects.

Plugging n, k into the the formula:

[tex]C(10, 7) = \frac{10!}{7! \cdot 3!} = \frac{10 \cdot 9 \cdot 8 \cdot 7!}{7! \ cdot 3!} = 10 \cdot 9 \cdot 8 = 120[/tex]

Then I conclude that for every combination of the first 3 digits there exist 120 ways of rearranging the last 7 digits.

So the total amount of phone numbers which can be written in such a manner is:

Total ways = 12 * 120 = 1440

I don't know if the answer is truly correct, but I think intuitively it makes sense.