Key Concepts and Formulas

• Combinations: The number of ways to choose $k$ objects from a set of $n$ distinct objects is given by $\binom{n}{k} = \frac{n!}{k!(n-k)!}$.
• Permutations of a Multiset: The number of ways to arrange $n$ objects, where $n_1$ are of one kind, $n_2$ are of another kind, ..., $n_k$ are of the $k$-th kind, is given by $\frac{n!}{n_1!n_2!...n_k!}$.
• Casework: Dividing a problem into mutually exclusive cases and summing the results.

Step-by-Step Solution

Step 1: Analyze the Letters in 'EXAMINATION'

We need to identify the distinct letters and their frequencies in the word 'EXAMINATION'. This is crucial for determining the possible cases when forming 4-letter words.

• E: 1
• X: 1
• A: 2
• M: 1
• I: 2
• N: 2
• T: 1
• O: 1

Therefore, we have 8 distinct letters (E, X, A, M, I, N, T, O), with A, I, and N appearing twice each, and the rest appearing once.

Step 2: Casework for Forming 4-Letter Words

We will now consider all possible cases for forming 4-letter words based on the repetition of letters.

Case 1: All four letters are distinct.

We need to choose 4 distinct letters from the 8 available distinct letters.

• Selection: The number of ways to choose 4 distinct letters from 8 is $\binom{8}{4}$. $\binom{8}{4} = \frac{8!}{4!4!} = \frac{8 \times 7 \times 6 \times 5}{4 \times 3 \times 2 \times 1} = 70$
• Arrangement: The 4 distinct letters can be arranged in $4!$ ways. $4! = 4 \times 3 \times 2 \times 1 = 24$
• Total for Case 1: The total number of 4-letter words with distinct letters is $\binom{8}{4} \times 4! = 70 \times 24 = 1680$.

Case 2: Two letters are identical, and the other two are distinct.

We need to choose one letter to repeat twice, and then two distinct letters from the remaining letters.

• Selection of the repeated pair: We have 3 choices for the letter that appears twice (A, I, or N). So, $\binom{3}{1} = 3$.
• Selection of the two distinct letters: After choosing one repeated letter, we have 7 remaining distinct letters to choose from. We need to choose 2 distinct letters from these 7. So, $\binom{7}{2} = \frac{7 \times 6}{2 \times 1} = 21$.
• Arrangement: We have 4 letters, with one letter repeated twice. The number of arrangements is $\frac{4!}{2!} = \frac{24}{2} = 12$.
• Total for Case 2: The total number of 4-letter words with two identical letters and two distinct letters is $\binom{3}{1} \times \binom{7}{2} \times \frac{4!}{2!} = 3 \times 21 \times 12 = 756$.

Case 3: Two letters are identical, and another two letters are identical.

We need to choose two letters to repeat twice each.

• Selection of the two pairs: We have 3 letters that appear twice (A, I, N). We need to choose 2 of them. So, $\binom{3}{2} = \frac{3 \times 2}{2 \times 1} = 3$.
• Arrangement: We have 4 letters, with two letters repeated twice each. The number of arrangements is $\frac{4!}{2!2!} = \frac{24}{2 \times 2} = 6$.
• Total for Case 3: The total number of 4-letter words with two pairs of identical letters is $\binom{3}{2} \times \frac{4!}{2!2!} = 3 \times 6 = 18$.

Step 3: Summing Up All Valid Cases

The total number of 4-letter words is the sum of the number of words in each case.

Total number of words = Case 1 + Case 2 + Case 3 = $1680 + 756 + 18 = 2454$.

Common Mistakes & Tips

• Missing Cases: Ensure all possible scenarios are considered. A systematic approach based on the number of distinct letters is helpful.
• Incorrect Permutation Formula: Using the wrong formula for arranging letters with repetitions. Remember to divide by the factorial of the count of each repeated letter.
• Double Counting: Be careful not to double count when choosing letters, especially when dealing with letters that can be repeated.

Summary

By analyzing the frequencies of the letters in 'EXAMINATION' and considering different cases based on letter repetitions, we calculated the number of 4-letter words that can be formed. We found 1680 words with all distinct letters, 756 words with one pair of identical letters and two distinct letters, and 18 words with two pairs of identical letters. Summing these up, we get a total of 2454 possible 4-letter words.

The final answer is $\boxed{2454}$.