How many different words can be made with the letter of the wor NAINITAL such that each of the word begins with L and end with T ?

Options

A. 85
B. 88
C. 92
D. 90

Answer

To solve the problem of how many different words can be formed using the letters of the word "NAINITAL" such that each word begins with 'L' and ends with 'T', we can follow these steps:

### Step-by-Step Solution:

1. **Identify the Fixed Letters**:
Since the word must start with 'L' and end with 'T', we can fix these two letters. This gives us the structure:
\[ L \_ \_ \_ \_ \_ T \]

2. **Count the Remaining Letters**:
After fixing 'L' and 'T', we need to arrange the remaining letters from "NAINITAL". The letters left are: N, A, I, N, I, A (since we have removed one 'L' and one 'T').

3. **List the Remaining Letters**:
The remaining letters are:
- N: 2
- A: 2
- I: 2

4. **Calculate the Total Arrangements**:
The total number of arrangements of the remaining letters (N, A, I, N, I, A) can be calculated using the formula for permutations of multiset:
\[ \text{Total arrangements} = \frac{n!}{n_1! \cdot n_2! \cdot n_3!} \]
where \( n \) is the total number of letters to arrange, and \( n_1, n_2, n_3 \) are the frequencies of the distinct letters.

Here, we have:
- Total letters to arrange = 6 (N, A, I, N, I, A)
- Frequencies: N = 2, A = 2, I = 2

Thus, the formula becomes:
\[ \text{Total arrangements} = \frac{6!}{2! \cdot 2! \cdot 2!} \]

5. **Calculate Factorials**:
- \( 6! = 720 \)
- \( 2! = 2 \)

Therefore:
\[ \text{Total arrangements} = \frac{720}{2 \cdot 2 \cdot 2} = \frac{720}{8} = 90 \]

6. **Conclusion**:
The total number of different words that can be formed under the given conditions is **90**.

### Final Answer:
The number of different words that can be formed is **90**.