To solve the problem of finding the number of 4-letter words that can be formed from the letters of the word "EXAMINATION," we will break down the solution into several cases based on the repetition of letters.
### Step 1: Identify the 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
This gives us a total of 11 letters, with some letters repeating.
### Step 2: Count the distinct letters
The distinct letters in "EXAMINATION" are: E, X, A, M, I, N, T, O.
Thus, we have 8 distinct letters.
### Step 3: Case 1 - All letters are distinct
In this case, we choose 4 letters from the 8 distinct letters. The number of ways to choose 4 letters from 8 is given by the combination formula \( \binom{n}{r} \).
\[
\text{Ways to choose 4 letters} = \binom{8}{4}
\]
After choosing the letters, we can arrange them in \( 4! \) ways.
\[
\text{Total for Case 1} = \binom{8}{4} \times 4!
\]
Calculating \( \binom{8}{4} \):
\[
\binom{8}{4} = \frac{8!}{4!(8-4)!} = \frac{8 \times 7 \times 6 \times 5}{4 \times 3 \times 2 \times 1} = 70
\]
Now calculating \( 4! \):
\[
4! = 24
\]
Thus, the total for Case 1 is:
\[
70 \times 24 = 1680
\]
### Step 4: Case 2 - One letter is repeated, and two are different
Here, we can have two scenarios:
1. The letter 'A' is repeated.
2. The letter 'I' is repeated.
3. The letter 'N' is repeated.
For each of these scenarios, we will choose 1 letter from the repeating letters and 2 from the remaining distinct letters.
#### Sub-case 2.1: Choosing 'A' as the repeated letter
- Choose 2 letters from the remaining 7 distinct letters (E, X, M, I, N, T, O):
\[
\text{Ways to choose 2 letters} = \binom{7}{2}
\]
The arrangement of these letters (A, A, and the two chosen letters) will be:
\[
\frac{4!}{2!} \quad \text{(since 'A' is repeated)}
\]
Calculating:
\[
\binom{7}{2} = \frac{7 \times 6}{2 \times 1} = 21
\]
Thus, the total for this sub-case is:
\[
21 \times \frac{4!}{2!} = 21 \times \frac{24}{2} = 21 \times 12 = 252
\]
#### Sub-case 2.2: Choosing 'I' as the repeated letter
Following the same logic:
\[
\text{Total for this sub-case} = 252
\]
#### Sub-case 2.3: Choosing 'N' as the repeated letter
Again, the same calculation applies:
\[
\text{Total for this sub-case} = 252
\]
Adding these sub-cases together for Case 2:
\[
252 + 252 + 252 = 756
\]
### Step 5: Case 3 - Two letters are alike
Here we can have:
1. Two 'A's and two 'I's.
2. Two 'A's and two 'N's.
3. Two 'I's and two 'N's.
For each of these scenarios, we will choose 2 letters from the repeating letters.
#### Sub-case 3.1: Choosing 'A' and 'I'
The arrangement of these letters (A, A, I, I) will be:
\[
\frac{4!}{2! \times 2!} = \frac{24}{4} = 6
\]
#### Sub-case 3.2: Choosing 'A' and 'N'
The arrangement will also be:
\[
\frac{4!}{2! \times 2!} = 6
\]
#### Sub-case 3.3: Choosing 'I' and 'N'
The arrangement will also be:
\[
\frac{4!}{2! \times 2!} = 6
\]
Adding these sub-cases together for Case 3:
\[
6 + 6 + 6 = 18
\]
### Step 6: Total number of ways
Now, we can sum the totals from all cases:
\[
\text{Total} = \text{Case 1} + \text{Case 2} + \text{Case 3} = 1680 + 756 + 18 = 2454
\]
### Final Answer
The total number of 4-letter words that can be formed from the letters of the word "EXAMINATION" is **2454**.
