The number of words which can be formed out of the letters of the word ALLAHABAD such that the vowels occupy the even positions is

Options

A. 60
B. 90
C. 120
D. none of these

Answer

To solve the problem of forming words from the letters of "ALLAHABAD" such that the vowels occupy the even positions, we can follow these steps:

### Step 1: Identify the letters and their counts
The word "ALLAHABAD" consists of the following letters:
- A (4 times)
- L (2 times)
- H (1 time)
- B (1 time)
- D (1 time)

### Step 2: Determine the positions
The total number of letters is 9. The even positions in a 9-letter word are positions 2, 4, 6, and 8. Therefore, there are 4 even positions available for the vowels.

### Step 3: Place the vowels in the even positions
The only vowel available is 'A', which appears 4 times. Since we have 4 even positions and 4 'A's, we will place one 'A' in each of the even positions.

### Step 4: Determine the remaining letters
After placing the vowels, we have the following letters left to arrange:
- L (2 times)
- H (1 time)
- B (1 time)
- D (1 time)

This gives us a total of 5 letters (L, L, H, B, D) to arrange in the remaining odd positions (1, 3, 5, 7, 9).

### Step 5: Calculate the arrangements of the remaining letters
The number of ways to arrange the remaining letters (L, L, H, B, D) can be calculated using the formula for permutations of multiset:

\[ \text{Number of arrangements} = \frac{n!}{n_1! \cdot n_2! \cdots n_k!} \]

Where:
- \( n \) = total number of letters to arrange (5 in this case)
- \( n_1, n_2, \ldots \) = counts of each distinct letter

In our case:
- Total letters (n) = 5
- L appears 2 times, while H, B, and D appear 1 time each.

Thus, the formula becomes:

\[ \text{Number of arrangements} = \frac{5!}{2! \cdot 1! \cdot 1! \cdot 1!} \]

### Step 6: Calculate the factorial values
Calculating the factorials:
- \( 5! = 120 \)
- \( 2! = 2 \)
- \( 1! = 1 \)

Now substituting these values into the formula:

\[ \text{Number of arrangements} = \frac{120}{2 \cdot 1 \cdot 1 \cdot 1} = \frac{120}{2} = 60 \]

### Final Answer
Thus, the total number of words that can be formed from the letters of "ALLAHABAD" such that the vowels occupy the even positions is **60**.

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