# Is this correct?

consider determining how many possible phone numbers are in an area code (repeated numbers allowed)

Is this a combination, a permutation, or neither?

I think it would be a combination wouldn't it?

Since there are ten number possibilities in each of the three spaces of the area code and repeats are allowed, there are 10^3 (1000) possible permutations.

To distinguish combinations from permutations, see the following sources:

http://mathforum.org/dr.math/faq/faq.comb.perm.html

http://tutors4you.com/permutationcombinationtutorial.htm

http://en.wikipedia.org/wiki/Combinatorics

http://betterexplained.com/articles/easy-permutations-and-combinations/

I hope this helps. Thanks for asking.

## You are correct, it would be a combination.

To determine how many possible phone numbers are in an area code, with repeated numbers allowed, we need to consider the number of choices for each digit of the phone number. In this case, there are 10 number possibilities (0-9) for each of the three spaces of the area code.

To find the total number of possible combinations, we use the formula for combinations, which is n^r, where n is the number of choices for each digit (10 in this case) and r is the number of digits (3 in this case).

So the total number of possible phone numbers would be 10^3, which is equal to 1000.

It's important to note that combinations are used when the order of the elements does not matter, while permutations are used when the order does matter. In this case, the order of the numbers in the area code does not matter, so it would be a combination.