I'm taking a discrete math course in university right now and we're studying permutations and combinations as one of the chapters. The question I'm confused about states that given a 5-letter alphabet, {A, B, C, D, E}, how many 3-letter strings can be made? Here's my attempt:
Case 1: no duplicates. This is simply asking how many 3-letter permutations exist over the set, so we have $\frac{5!}{(5-3)!}=(5)(4)(3)$. This intuitively makes sense too, because for the first letter, we have 5 choices, for the 2nd we have 4 choices, and for the 3rd we have 3 choices, so using the multiplication principle we should get $(5)(4)(3)$.
Case 2: 1 duplicate. Here, I claim that we have 5 choices for the first letter, but then that the second letter must be a duplicate of the first, so we only have 1 choice for it. Lastly, the third letter can be any of the remaining letters except the duplicated one, so there are 4 choices. This gives the specific permutation that the duplicated letter is in positions 1 and 2, but it could also be in positions 1 and 3 or 2 and 3, so we need to multiply our previous answer by 3. This gives: $((5)(1)(4))(3)=60$
Case 3: 2 duplicates. Finally, I claim that to have 2 duplicates, there is only one case, in which we have $(5)(1)(1)$ choices, and any other orderings would just give the same string.
Adding up all the cases, we get $60+60+5=125$ strings.
However, when I look at the solution, the book's answer is much more simple. They simply state that we allow repeated letters, so we can just do $(5)(5)(5)$. I was wondering what is wrong with my reasoning?