The alphabets of word ALLAHABAD are arranged at random. The probability that in the words so formed, all identical alphabets are found together, is

Options

A. `1//63`
B. `16//17`
C. `5!//9!`
D. None of these

Answer

To solve the problem, we need to find the probability that in the arrangements of the letters of the word "ALLAHABAD", all identical alphabets are found together.

### Step-by-Step Solution:

1. **Identify the letters and their frequencies**:
The word "ALLAHABAD" consists of the following letters:
- A: 5 times
- L: 2 times
- H: 1 time
- B: 1 time
- D: 1 time

2. **Total number of letters**:
The total number of letters in "ALLAHABAD" is 9.

3. **Calculate the total arrangements of the letters**:
The total arrangements of the letters can be calculated using the formula for permutations of multiset:
\[ \text{Total arrangements} = \frac{n!}{n_1! \times n_2! \times n_3! \times \ldots} \]
Here, \(n\) is the total number of letters, and \(n_1, n_2, \ldots\) are the frequencies of each letter.
\[ \text{Total arrangements} = \frac{9!}{5! \times 2! \times 1! \times 1! \times 1!} \]

4. **Calculate the arrangements where identical letters are together**:
To find the arrangements where all identical letters are together, we can treat all identical letters as a single unit.
- Group the 5 A's together as one unit (let's call it AAAAA).
- The remaining letters are L, L, H, B, D.
- This gives us the units: AAAAA, L, L, H, B, D (which is a total of 6 units).

5. **Calculate the arrangements of these units**:
The arrangements of these 6 units (where L is repeated) can be calculated as:
\[ \text{Favorable arrangements} = \frac{6!}{2!} \]

6. **Calculate the probability**:
The probability that all identical letters are together is given by the ratio of favorable arrangements to total arrangements:
\[ P(\text{all identical together}) = \frac{\text{Favorable arrangements}}{\text{Total arrangements}} = \frac{\frac{6!}{2!}}{\frac{9!}{5! \times 2! \times 1! \times 1! \times 1!}} \]

7. **Simplify the expression**:
\[ P = \frac{6! \times 5! \times 2! \times 1! \times 1! \times 1!}{2! \times 9!} \]
\[ = \frac{720 \times 120}{2 \times 362880} = \frac{86400}{725760} = \frac{1}{8.3333} \approx \frac{1}{63} \]

### Final Answer:
The probability that in the words formed, all identical alphabets are found together is \(\frac{1}{63}\).

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