Hello, I need help, I cant remember my cell phone pin i set last night. It is a four digit number using numbers 0-9, the only thing i remember is it does contain 1 and 3. How many possible combinations could there possibly be? I don't want to reset it, I will lose all my information.

Hello!
There are 10^4 pins at all. Denote the set of pins that contain no 1's as B_1 and that contain no 3's as B_3.
We'll use the formula n(A uu B) = n(A)+n(B)-n(A nn B) which is true for any finite sets, particularly for B_1 and B_3 (n(()) means the number of elements).
The set of pins that contain 1 AND contain 3 is the complement of those which don't contain 1 OR don't contain 3, i.e. (B_1 uu B_3)^C. The number of pins in this set is 10^4 - n(B_1 uu B_3). By the above formula it is
10^4 - (n(B_1)+n(B_3)-n(B_1 nn B_3)).
It is clear that n(B_1)=9^4 (any number except 1 at any of 4 positions), and n(B_3) is the same. n(B_1 nn B_3) is the number of pins that don't contain 1 AND don't contain 3, there are 8^4 of those (any number except 1 and 3 at any position).
So the answer is 10^4 - (9^4+9^4-8^4)=10^4+8^4-2*9^4 = 974.
There are too many variants to try, it is necessary to recall the pin. Try the following trick: imagine that you want to set a new pin, what could it be?