Problem Solving

If you are creating 1 Letter Code you can get 26 codes
Two letter codes with repetition allowed = 26 x 26 = 576
Three Letter codes = 26 x 26 x 26 = 26^3

You know that any power of 6 ends in a 6
And the Units digit of all 3 numbers is a 6
Which means your answer must end in an 8

Let us start by spelling out the rule of counting:
The rule of counting says that if we have m ways of selecting first item, n ways of selecting second item, o ways of selecting third item and so on then we have a total of m*n*o*... total ways of selecting items.

The second thing that we should be aware of in (probability) problems of these types is whether the problem calls for "or" or "and" operation. In probability theory when you see the word "or" it is an indication for you to use "+" and when you see "and" you apply "*".

This problem calls for "or" indicating us to sum up the numbers.

So far so good. Now let us get into the math.

1. Three letter coded stock: Imagine three placeholders _, _, _. Now the first place can be filled with any alphabets from A to Z in 26 ways. Also since we are free to repeat the alphabets, we would have another 26 ways of filling the second placeholder, and continuing in this fashion, we can fill the third place in another 26 ways. So the total numbers of choices we have for three letter coded stock is 26*26*26 = 17,576

2. Two letter coded stock: Here we have two placeholders _,_. Extending the logic from point 1 above, we would have 26*26 = 676 choices.

3. One letter coded stock: Here we have one placeholder _, and so only 26 choices.

Final Answer: Now since the problem read "...one-, two-, or, three...", we would need to sum (+) up the numbers i.e. 17,576+676+26 = 18,278 choices.

So "E" is the correct answer choice.