How many different words can be formed with the letters of the word NAINITAL such that each of the word begin with L and end with T?

Options

A. a. 90
B. b. 80
C. c. 88
D. d. 82

Answer

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

### Step 1: Identify the letters in "NAINITAL"
The word "NAINITAL" consists of the following letters:
- N (2 times)
- A (2 times)
- I (2 times)
- T (1 time)
- L (1 time)

### Step 2: Fix the positions of 'L' and 'T'
Since we want the words to start with 'L' and end with 'T', we can fix these letters in their respective positions:
- First position: L
- Last position: T

This leaves us with the letters in between to arrange:
- Remaining letters: N, A, I, N, I, A (total of 6 letters)

### Step 3: Count the arrangements of the remaining letters
Now we need to find the number of distinct arrangements of the remaining letters (N, A, I, N, I, A).

The formula for the number of arrangements of a multiset is given by:
\[ \text{Number of arrangements} = \frac{n!}{n_1! \cdot n_2! \cdot n_3! \cdots} \]
Where:
- \( n \) is the total number of items,
- \( n_1, n_2, n_3, \ldots \) are the frequencies of the distinct items.

In our case:
- Total letters (n) = 6 (N, A, I, N, I, A)
- Frequencies:
- N appears 2 times
- A appears 2 times
- I appears 2 times

Thus, we can calculate the arrangements as follows:
\[ \text{Number of arrangements} = \frac{6!}{2! \cdot 2! \cdot 2!} \]

### Step 4: Calculate the factorials
Now, we calculate the factorials:
- \( 6! = 720 \)
- \( 2! = 2 \)

So we have:
\[ \text{Number of arrangements} = \frac{720}{2 \cdot 2 \cdot 2} = \frac{720}{8} = 90 \]

### Final Answer
Thus, the total number of different words that can be formed with the letters of the word "NAINITAL" that begin with 'L' and end with 'T' is **90**.

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