If the different permutations of the word EXAMINATION are listed as in a dictinary how many items are there in this list before the first word starting with E?

Options

A. `906200`
B. `907200`
C. `908200`
D. None of these

Answer

To solve the problem of finding how many permutations of the word "EXAMINATION" come before the first word starting with 'E', we can follow these steps:

### Step 1: Identify the letters in "EXAMINATION"
The word "EXAMINATION" consists of the following letters:
- E
- X
- A
- M
- I (occurs twice)
- N (occurs twice)
- T
- O

### Step 2: Count the total letters and their frequencies
The total number of letters in "EXAMINATION" is 11. The frequency of each letter is:
- E: 1
- X: 1
- A: 1
- M: 1
- I: 2
- N: 2
- T: 1
- O: 1

### Step 3: Calculate permutations starting with letters before 'E'
In alphabetical order, the letters before 'E' in "EXAMINATION" are 'A', 'I', 'M', 'N', 'O', and 'X'. We will calculate the permutations starting with each of these letters.

#### Case 1: Starting with 'A'
If 'A' is fixed at the first position, we have the remaining letters: E, X, M, I, I, N, N, T, O (10 letters total).
The number of permutations is calculated as:
\[ \frac{10!}{2! \times 2!} = \frac{3628800}{4} = 907200 \]

#### Case 2: Starting with 'I'
If 'I' is fixed at the first position, we have the remaining letters: E, X, A, M, I, N, N, T, O (10 letters total).
The number of permutations is:
\[ \frac{10!}{2! \times 2!} = 907200 \]

#### Case 3: Starting with 'M'
If 'M' is fixed at the first position, we have the remaining letters: E, X, A, I, I, N, N, T, O (10 letters total).
The number of permutations is:
\[ \frac{10!}{2! \times 2!} = 907200 \]

#### Case 4: Starting with 'N'
If 'N' is fixed at the first position, we have the remaining letters: E, X, A, I, I, N, T, O (10 letters total).
The number of permutations is:
\[ \frac{10!}{2! \times 2!} = 907200 \]

#### Case 5: Starting with 'O'
If 'O' is fixed at the first position, we have the remaining letters: E, X, A, I, I, N, N, T (10 letters total).
The number of permutations is:
\[ \frac{10!}{2! \times 2!} = 907200 \]

#### Case 6: Starting with 'X'
If 'X' is fixed at the first position, we have the remaining letters: E, A, I, I, N, N, T, O (10 letters total).
The number of permutations is:
\[ \frac{10!}{2! \times 2!} = 907200 \]

### Step 4: Sum the permutations
Now, we sum all the permutations calculated for each case:
\[ 907200 (A) + 907200 (I) + 907200 (M) + 907200 (N) + 907200 (O) + 907200 (X) = 6 \times 907200 = 5443200 \]

### Conclusion
The total number of permutations of the word "EXAMINATION" that come before the first word starting with 'E' is **5443200**.