To solve the problem of how many four-letter words can be formed using the letters of the word 'INEFFECTIVE', we will consider different cases based on the repetition of letters.
### Step-by-Step Solution:
1. **Identify the Letters and Their Frequencies**:
The letters in 'INEFFECTIVE' are:
- I: 2
- N: 1
- E: 3
- F: 2
- C: 1
- T: 1
- V: 1
Total unique letters = 7 (I, N, E, F, C, T, V).
2. **Case 1: All Different Letters (abcd type)**:
- We need to choose 4 letters from the 7 unique letters.
- The number of ways to choose 4 letters from 7 is given by \( \binom{7}{4} \).
- The arrangements of these 4 letters is \( 4! \).
- Therefore, the total for this case is:
\[
\text{Total} = \binom{7}{4} \times 4! = 35 \times 24 = 840
\]
3. **Case 2: Two Letters Same, Two Different (aabc type)**:
- We can have two of one letter and two different letters.
- The only letters that can be repeated are I, E, or F.
- Choose 1 letter to be repeated (from I, E, F) and then choose 2 different letters from the remaining letters.
- The number of ways to choose 1 letter from {I, E, F} is \( \binom{3}{1} = 3 \).
- The remaining letters after choosing one to repeat are 6 (since we have 1 letter already chosen).
- The number of ways to choose 2 letters from these 6 is \( \binom{6}{2} \).
- The arrangements of these letters (where 2 are the same) is \( \frac{4!}{2!} \).
- Therefore, the total for this case is:
\[
\text{Total} = \binom{3}{1} \times \binom{6}{2} \times \frac{4!}{2!} = 3 \times 15 \times 12 = 540
\]
4. **Case 3: One Letter Different, Three Same (aaab type)**:
- The only letter that can be used three times is E.
- Choose 1 letter from the remaining letters (I, N, F, C, T, V) to be different.
- The number of ways to choose 1 letter from 6 is \( \binom{6}{1} = 6 \).
- The arrangements of these letters (where 3 are the same) is \( \frac{4!}{3!} \).
- Therefore, the total for this case is:
\[
\text{Total} = \binom{6}{1} \times \frac{4!}{3!} = 6 \times 4 = 24
\]
5. **Case 4: Two Letters Same, Two Letters Same (aabb type)**:
- We can have two of one letter and two of another letter.
- The possible pairs are (I, E), (I, F), (E, F).
- The number of ways to choose 2 letters from {I, E, F} is \( \binom{3}{2} = 3 \).
- The arrangements of these letters (where 2 are the same and another 2 are the same) is \( \frac{4!}{2!2!} \).
- Therefore, the total for this case is:
\[
\text{Total} = \binom{3}{2} \times \frac{4!}{2!2!} = 3 \times 6 = 18
\]
6. **Final Calculation**:
- Now, we sum the totals from all cases:
\[
\text{Total Words} = 840 + 540 + 24 + 18 = 1422
\]
### Conclusion:
The total number of four-letter words that can be formed using the letters of the word 'INEFFECTIVE' is **1422**.
