PIN number combination that has a 3 and 8 in it. So the question is, How many PINs(4 numbers from 0 to 9 with repeat) are there when 3 and 8 (in unknown positions) are in that pin and also the remaining 2 positions can be any number between 0 to 9.
My solution:
Number of combinations without the 3 and 8 in them=> 84 = 4096
then i subtract 104 from 4096 which gives 5904 combinations WITH a 3 AND 8 in them. 
Is the result correct? 
 A: No, your result is not correct.
4096 is how many PINs there are with neither 3 nor 8 in them, so some of your 5904 combinations will be ones like 9939 that don't satisfy your conditions.
However, the computation suggested by Markos Karamenes:

Look at the problem this way: you have $10^2$ combinations from the two unknown numbers and 3 possible positions for the 3,8. Can you count them now?

doesn't work either, even if we correct "3 possible positions" to 12. It leads to overcounting of combinations such as 3883.
You need to amend your method by subtracting all of the combinations that have a 3 but not an 8, or vice versa. (There are $9^4-8^4$ of each of these possibilities -- why?)
A: For $i\in\{0,1,\dots,9\}$ let $P_i$ be the set of pins that do not contain digit $i$.
Then to be found is: $$|P_3^{\complement}\cap P_8^{\complement}|=|(P_3\cup P_8)^{\complement}|=10^4-|P_3\cup P_8|$$ 
Applying the principle of inclusion/exclusion we find:$$|P_3^{\complement}\cap P_8^{\complement}|=10^4-|P_3|-|P_8|+|P_3\cap P_8|=10^4-9^4-9^4+8^4=974$$
