How many 3 digit area codes can be formed using 2,5,7and 9 (with no repeating digits? What is the probability that one of these codes can be divisible by 9 but not divisible by 2?

Okay, the total number of possible codes are when you choose 3 of the 4 numbers, and order is important here. Can you get that? 'choose' mean you pick in any order (using combinations), and then, you need to arrange them.

Either that way of thinking, or you find the number of ways directly, using permutations.

Then, you only have to look for the number of codes which are divisible by nine and which are odd. Meaning the number should end with 5, 7 or 9 and the sum of all the digits should give a multiple of 9.

For example, 729 is divisible by 9 since 7+2+9 = 18 and 18 is a multiple of 9.

Can you give your problem a try? :)