Remember that you are counting the number of items in some list of outcomes. Write down part of this list. Write down an element in the middle of the list – how are you deciding whether your element really is in the list. Could you get this element more than once using your proposed answer?
If generating an element on the list involves selecting something (for example, picking a letter or picking a position to put a letter, etc), can the things you select be repeated? Remember, permutations and combinations select objects from a set without repeats.
Does order matter? Be careful here and be sure you know what your answer really means. We usually say that order matters when you get different outcomes when the same objects are selected in different orders. Combinations and “Stars & Bars” are used when order does not matter.
There are four possibilities when it comes to order and repeats. If order matters and repeats are allowed, the answer will look like \(n^k\text{.}\) If order matters and repeats are not allowed, we have \(P(n,k)\text{.}\) If order doesn't matter and repeats are allowed, use stars and bars. If order doesn't matter and repeats are not allowed, use \({n\choose k}\text{.}\) But be careful: this only applies when you are selecting things, and you should make sure you know exactly what you are selecting before determining which case you are in.
Think about how you would represent your counting problem in terms of sets or functions. We know how to count different sorts of sets and different types of functions.
As we saw with combinatorial proofs, you can often solve a counting problem in more than one way. Do that, and compare your numerical answers. If they don't match, something is amiss.