The total number of permutations of 4 letters that can be made out of the letters of the word EXAMINATION is

Options

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

Answer

To find the total number of permutations of 4 letters that can be made from the letters of the word "EXAMINATION", we will follow these steps:

### Step 1: Identify the letters and their frequencies
The word "EXAMINATION" consists of 11 letters in total. The breakdown of the letters is as follows:
- E: 1
- X: 1
- A: 1
- M: 1
- I: 2
- N: 2
- T: 1
- O: 1

### Step 2: Determine the total number of letters and repetitions
From the breakdown, we see that:
- Total letters = 11
- Repeated letters: I (2 times), N (2 times)

### Step 3: Calculate the number of permutations based on different cases of letter selection
We will consider different cases based on the selection of letters:

#### Case 1: All 4 letters are different
We can select 4 different letters from the 8 unique letters (E, X, A, M, I, N, T, O).
- The number of ways to choose 4 letters from 8 = C(8, 4) = 70
- The number of permutations of 4 different letters = 4! = 24
- Total permutations for this case = 70 * 24 = 1680

#### Case 2: 3 letters are different and 1 letter is repeated
The only letters that can be repeated are I and N. We will consider each case separately.

- **Sub-case 2.1: One letter is I (3 different letters + I)**
- Choose 3 from the remaining 7 letters (E, X, A, M, N, T, O).
- The number of ways to choose 3 letters = C(7, 3) = 35
- The number of permutations = 4! / 1! = 24
- Total permutations for this sub-case = 35 * 24 = 840

- **Sub-case 2.2: One letter is N (3 different letters + N)**
- Choose 3 from the remaining 7 letters (E, X, A, M, I, T, O).
- The number of ways to choose 3 letters = C(7, 3) = 35
- The number of permutations = 4! / 1! = 24
- Total permutations for this sub-case = 35 * 24 = 840

Total permutations for Case 2 = 840 + 840 = 1680

#### Case 3: 2 letters are I and 2 letters are different
We can choose 2 different letters from the remaining 7 letters (E, X, A, M, N, T, O).
- The number of ways to choose 2 letters = C(7, 2) = 21
- The number of permutations = 4! / 2! = 12
- Total permutations for this case = 21 * 12 = 252

#### Case 4: 2 letters are N and 2 letters are different
Similar to Case 3, we can choose 2 different letters from the remaining 7 letters (E, X, A, M, I, T, O).
- The number of ways to choose 2 letters = C(7, 2) = 21
- The number of permutations = 4! / 2! = 12
- Total permutations for this case = 21 * 12 = 252

#### Case 5: 2 letters are I and 2 letters are N
- The number of permutations = 4! / (2! * 2!) = 6

### Step 4: Sum up all the cases
Total permutations = Case 1 + Case 2 + Case 3 + Case 4 + Case 5
= 1680 + 1680 + 252 + 252 + 6
= 3870

### Final Answer
The total number of permutations of 4 letters that can be made out of the letters of the word "EXAMINATION" is **3870**.

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