How many different words can be formed by using all the letters of the word ALLAHABAD?

Options

A. 30
B. (9!)
C. (9!)/(2!)
D. (9!)/(2! 4!)

Explanation

To find the number of different words that can be formed using all the letters of the word "ALLAHABAD", we can follow these steps:

### Step 1: Count the total number of letters
The word "ALLAHABAD" consists of 9 letters.

### Step 2: Identify the frequency of each letter
In the word "ALLAHABAD":
- A appears 4 times
- L appears 2 times
- H appears 1 time
- B appears 1 time
- D appears 1 time

### Step 3: Use the formula for permutations of multiset
The formula to find the number of distinct permutations of a set of items where there are repetitions is given by:

\[ \text{Number of arrangements} = \frac{n!}{p_1! \times p_2! \times \ldots \times p_k!} \]

Where:
- \( n \) is the total number of items,
- \( p_1, p_2, \ldots, p_k \) are the frequencies of the repeated items.

### Step 4: Apply the formula
In our case:
- \( n = 9 \) (total letters)
- \( p_1 = 4 \) (for A)
- \( p_2 = 2 \) (for L)
- \( p_3 = 1 \) (for H)
- \( p_4 = 1 \) (for B)
- \( p_5 = 1 \) (for D)

Thus, the number of different arrangements is:

\[ \text{Number of arrangements} = \frac{9!}{4! \times 2! \times 1! \times 1! \times 1!} \]

### Step 5: Calculate the factorials
Now we calculate the factorials:
- \( 9! = 362880 \)
- \( 4! = 24 \)
- \( 2! = 2 \)
- \( 1! = 1 \) (for H, B, and D)

### Step 6: Substitute the values into the formula
Now substituting the values into the formula:

\[ \text{Number of arrangements} = \frac{362880}{24 \times 2 \times 1 \times 1 \times 1} = \frac{362880}{48} = 7560 \]

### Final Answer
Therefore, the total number of different words that can be formed using all the letters of the word "ALLAHABAD" is **7560**.

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