The number of 4 letter words (with or without meaning) that can be formed from the letter of the work EXAMINATION is

Answer

To find the number of 4-letter words that can be formed from the letters of the word "EXAMINATION", we will analyze the distinct letters and their frequencies, and then consider different cases based on the repetition of letters.

### Step-by-Step Solution:

1. **Identify the distinct letters and their frequencies:**
The word "EXAMINATION" consists of the following letters:
- E: 1
- X: 1
- A: 2
- M: 1
- I: 2
- N: 2
- T: 1
- O: 1

Total distinct letters = 8 (E, X, A, M, I, N, T, O)

2. **Case Analysis:**
We will consider three cases based on the repetition of letters.

**Case 1: All letters are distinct.**
- We need to choose 4 distinct letters from 8.
- The number of ways to choose 4 letters from 8 is given by \( \binom{8}{4} \).
- The number of arrangements of 4 distinct letters is \( 4! \).
- Thus, the total for this case is:
\[ \text{Total for Case 1} = \binom{8}{4} \times 4! = 70 \times 24 = 1680 \]

3. **Case 2: Two letters are alike, and two letters are distinct.**
- The letters that can be alike are A, I, and N (since they appear twice).
- We can choose 1 letter from these 3 to be the repeated letter, which can be done in \( \binom{3}{1} \) ways.
- We then choose 2 distinct letters from the remaining 7 letters (after choosing the repeated letter).
- The number of ways to choose 2 letters from 7 is \( \binom{7}{2} \).
- The arrangement of the letters (2 alike and 2 distinct) is given by \( \frac{4!}{2!} \).
- Thus, the total for this case is:
\[ \text{Total for Case 2} = \binom{3}{1} \times \binom{7}{2} \times \frac{4!}{2!} = 3 \times 21 \times 12 = 756 \]

4. **Case 3: Two pairs of alike letters.**
- We can only form pairs from A, I, and N.
- We choose 2 letters from these 3 to be repeated, which can be done in \( \binom{3}{2} \) ways.
- The arrangement of the letters (2 of one kind and 2 of another) is given by \( \frac{4!}{2! \times 2!} \).
- Thus, the total for this case is:
\[ \text{Total for Case 3} = \binom{3}{2} \times \frac{4!}{2! \times 2!} = 3 \times 6 = 18 \]

5. **Total number of 4-letter words:**
- Now, we sum the totals from all cases:
\[ \text{Total} = \text{Total for Case 1} + \text{Total for Case 2} + \text{Total for Case 3} = 1680 + 756 + 18 = 2454 \]

### Final Answer:
The total number of 4-letter words that can be formed from the letters of the word "EXAMINATION" is **2454**.