Repeated Permutations In school we have been studying combinations and permutations, and in a programming assignment I was testing points on a coordinate plane. Testing all integer points surrounding the point (0,0) you obviously end up with 9 different permutations. 
(0,0) (0,1) (1,0) (1,1) (0,-1) (-1,0) (-1,-1) (1,-1) (-1,1)
I was trying to model this using the combination and permutation formulas we learned in class, and I arrive at an answer of 30 different permutations or 5 combinations. 
P = (6!/4!) = 30
C = 30/3! = 5
How would I take into account combinations that are not equivalent when reversed and disregard those that are? 
I apologize if this questions has been answered elsewhere, if nothing else I guess i'm asking what I should search to find what i'm looking for. 

I'd take the simpler approach of noting that there are 3 possibilities for each co-ordinate and 2 positions so the answer is $3\times3=9$ for a computational answer.
Now, if you want to look at them without order mattering, generally you'd break this down into cases and sum over those:
3 cases of the same digit repeated: (0,0), (1,1), (-1,-1)
If the digits don't repeat then it would be 3 choices for the first and 2 for the second value but there is a division by 2 because of double counting that leads to 3 possibilities: (0,1), (0,-1), and (1,-1).
The key is to consider whether order matters and is duplication allowed or not.
