How many different words of 4 letters can be formed with the letters of the word EXAMINATION?

Options

A. 2454
B. 2436
C. 2545
D. none of these

Answer

To solve the problem of how many different words of 4 letters can be formed with the letters of the word "EXAMINATION", we will break it down into cases based on the repetition of letters.

### Step-by-Step Solution:

1. **Identify the Letters and Their Frequencies**:
- The letters in "EXAMINATION" are:
- 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 1: All Letters are Distinct**:
- We need to choose 4 letters from the 8 distinct letters.
- The number of ways to choose 4 letters from 8 is given by \( \binom{8}{4} \).
- After choosing the letters, we can arrange them in \( 4! \) ways.
- Therefore, the total for this case is:
\[ \text{Total for Case 1} = \binom{8}{4} \times 4! = 70 \times 24 = 1680 \]

3. **Case 2: One Letter is Alike and Two are Distinct**:
- The letters that can be alike are A, I, or N (since they appear twice).
- We can choose 1 letter to be repeated (from A, I, N) in \( \binom{3}{1} \) ways.
- Then, we need to choose 2 more distinct letters from the remaining 7 letters.
- The number of ways to choose 2 letters from the remaining 7 is \( \binom{7}{2} \).
- The arrangement of these letters (where one letter is repeated) is \( \frac{4!}{2!} \).
- Therefore, 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 Letters are Alike**:
- We can choose 2 letters to be alike from A, I, or N (which appear twice).
- The number of ways to choose 2 letters from these 3 is \( \binom{3}{2} \).
- The arrangement of these letters (two pairs alike) is \( \frac{4!}{2! \times 2!} \).
- Therefore, 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. **Final Calculation**:
- Now, we add the totals from all cases:
\[ \text{Total Words} = \text{Case 1} + \text{Case 2} + \text{Case 3} = 1680 + 756 + 18 = 2454 \]

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