Nine different letters of the alphabet are given and five-letter words are formed using these letters. How many such five-letter words can be formed which contain at least one repeated letter?

Introduction / Context:
This question is about counting words (ordered strings of letters) allowing repetition and then restricting to those that actually have at least one repeated letter. Problems of this type are best handled using the complement method: first count all possible words, then subtract those with no repeated letters. This is a classic application of permutations with repetition and the principle of complementary counting in combinatorics.

Given Data / Assumptions:
- There are 9 distinct letters available.
- We are forming five-letter words (strings) using these letters.
- Repetition of letters is allowed in general, but we want only those words with at least one letter repeated.
- Order matters: two strings that differ in position of letters are considered different words.

Concept / Approach:
Directly counting all words that have at least one repeated letter is complicated, because many overlapping cases would need to be included. Instead, we use the complement approach. We first count all possible five-letter words that can be formed using the 9 letters with repetition allowed, and then subtract the number of words with no repeated letters at all (that is, all letters distinct). The result is the count of words where at least one letter appears more than once.

Step-by-Step Solution:
Step 1: Count the total number of five-letter words that can be formed from 9 letters when repetition is allowed.Step 2: For each of the 5 positions in the word, there are 9 choices (any of the 9 letters).Step 3: Total unrestricted words = 9^5 = 9 * 9 * 9 * 9 * 9 = 59049.Step 4: Next, count the number of five-letter words in which all letters are distinct (no repetition).Step 5: This is a permutation without repetition: 9P5 = 9 * 8 * 7 * 6 * 5.Step 6: Compute 9P5: 9 * 8 = 72; 72 * 7 = 504; 504 * 6 = 3024; 3024 * 5 = 15120.Step 7: Words with at least one repeated letter = total words - words with all distinct letters.Step 8: So required count = 59049 - 15120 = 43929.

Verification / Alternative check:
We can approximate the proportion of words with repeated letters by noting that 15120 is much smaller than 59049, so most words will indeed contain repetitions. Computing the difference gives 43929, and a quick check of the arithmetic 59049 - 15000 = 44049, then subtracting another 120 gives 43929, confirming the subtraction. This aligns with the logical expectation that only a small fraction of all possible five-letter words will have all distinct letters when repetition is allowed.

Why Other Options Are Wrong:
- 59049 counts all possible words, including those with no repetition; it ignores the condition of having at least one repeated letter.

- 15120 counts only words with all distinct letters, that is, the complement of what is required.

- 0 would suggest that no word has repetition, which is obviously incorrect when repetition is allowed in 9^5 possibilities.

Common Pitfalls:
A common mistake is to misinterpret the phrase "at least one letter repeated" and try to count cases of exactly one repetition, two repetitions, and so on separately, which becomes extremely complicated and prone to double-counting. The complement method efficiently avoids this complexity. Another error is confusing 9P5 with 9^5, or assuming that 9P5 already counts repeated letters, which it does not. Being clear on when repetition is allowed versus disallowed is crucial.

Final Answer:
The number of five-letter words that can be formed from 9 distinct letters with at least one repeated letter is 43929.