There are a series of of 2 letter 2 digit codes, how many codes can be created including repeated ...?

Question

.... letters or digits such as; 42cc,66qq,bb44, cc77?
Using numbers 0 through 9 and letters a through z

Answer 1

there are 26 letters and 10 digits

26*26 * 10*10 = 67600

But if the orders of the letters and digits do count, then
LLDD

4! / (2! * 2!) = 6

6 x 67600 = 405,600 ways

Answer 2

I hope I am not doing your homework for you.
Here is how to figure it out.
There are 10 digits (D) and 26 letters (L).
I you have one digit you have have ten possibilities. If you have two digits it is ten X ten or 100. That is half of your answer.
If you have one letter the possibilities are 26. With two letters it is 26 X 26 or ___ (do the math)
Since for every possibility of numbers you have all possibilities of numbers multiply the two results.
100 X ___
the formula is D X D X L X L.
three digits and 4 letters would be D X D X D X L X L X L X L
If you have more or less digits or numbers in a position say the first position could be a letter OR number that would be 10 + 26 (digits plus letters) in the first position. We will call that A and it is equal to 36.
Now it looks like A X D X L X L or 36 X 10 X 26 X 26

Answer 3

26^2 + 10^2 = 67,600
edit:
looking over your examples,
2(26^2 + 10^2) = 135,200