How many words can be formed from the all letters of the word INITIAL such that the all words must have started and ended with letter 'l' ?

Options

A. 120
B. 20
C. 5040
D. 240

Answer

To solve the problem of how many words can be formed from the letters of the word "INITIAL" such that all words must start and end with the letter 'L', we can follow these steps:

### Step-by-Step Solution:

1. **Identify the Letters**: The word "INITIAL" consists of the letters: I, N, I, T, I, A, L. The total number of letters is 7.

2. **Fix the Starting and Ending Letters**: Since we want the words to start and end with 'L', we will fix 'L' in the first and last positions. This means our arrangement looks like: L _ _ _ _ L.

3. **Count the Remaining Letters**: After fixing 'L' at both ends, we are left with the letters: I, N, I, T, I, A. This gives us a total of 5 letters to arrange in the middle.

4. **Identify Repeated Letters**: Among the remaining letters (I, N, I, T, I, A), we notice that the letter 'I' appears 3 times.

5. **Calculate the Arrangements**: The number of ways to arrange these 5 letters (I, N, I, T, I, A) can be calculated using the formula for permutations of multiset:
\[ \text{Number of arrangements} = \frac{n!}{p_1! \times p_2! \times \ldots} \]
where \( n \) is the total number of items to arrange, and \( p_1, p_2, \ldots \) are the frequencies of the repeated items.

Here, we have:
- Total letters (n) = 5
- The letter 'I' appears 3 times, and the other letters (N, T, A) appear once each.

Thus, the formula becomes:
\[ \text{Number of arrangements} = \frac{5!}{3! \times 1! \times 1!} \]

6. **Calculate Factorials**:
- \( 5! = 120 \)
- \( 3! = 6 \)
- \( 1! = 1 \)

Therefore:
\[ \text{Number of arrangements} = \frac{120}{6 \times 1 \times 1} = \frac{120}{6} = 20 \]

7. **Conclusion**: The total number of words that can be formed from the letters of the word "INITIAL" such that they start and end with 'L' is **20**.