All the letters of the word HAMSANANDI are placed at random in a row. The probability that the word ANAND occurs without getting split is

Options

A. `(1)/(42)`
B. `(1)/(60)`
C. `(1)/(420)`
D. `(1)/(329)`

Answer

To find the probability that the word "ANAND" occurs without getting split when all the letters of the word "HAMSANANDI" are arranged randomly, we can follow these steps:

### Step 1: Count the total number of arrangements of the letters in "HAMSANANDI".

The word "HAMSANANDI" consists of 10 letters in total, where:
- A appears 3 times,
- N appears 2 times,
- H, M, S, and D appear 1 time each.

The formula for the total arrangements of letters when there are repetitions is given by:

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

Where \( n \) is the total number of letters, and \( p_1, p_2, \ldots, p_k \) are the frequencies of the repeated letters.

Thus, we can calculate:

\[ \text{Total arrangements} = \frac{10!}{3! \times 2!} \]

### Step 2: Treat "ANAND" as a single entity.

To find the arrangements where "ANAND" occurs as a single entity, we can treat "ANAND" as one letter. Therefore, we have the following letters/entities to arrange:

- ANAND (as one entity),
- H,
- M,
- S,
- I.

This gives us a total of 6 entities to arrange.

### Step 3: Count the arrangements of these 6 entities.

The number of arrangements of these 6 entities is given by:

\[ \text{Arrangements with ANAND as one entity} = 6! \]

### Step 4: Calculate the probability.

The probability that "ANAND" occurs without getting split is given by the ratio of the number of favorable arrangements to the total arrangements:

\[ \text{Probability} = \frac{\text{Arrangements with ANAND as one entity}}{\text{Total arrangements}} \]

Substituting the values we calculated:

\[ \text{Probability} = \frac{6!}{\frac{10!}{3! \times 2!}} \]

### Step 5: Simplify the expression.

We can simplify this expression:

\[ \text{Probability} = \frac{6! \times 3! \times 2!}{10!} \]

Now, substituting \( 10! = 10 \times 9 \times 8 \times 7 \times 6! \):

\[ \text{Probability} = \frac{6! \times 3! \times 2!}{10 \times 9 \times 8 \times 7 \times 6!} \]

The \( 6! \) cancels out:

\[ \text{Probability} = \frac{3! \times 2!}{10 \times 9 \times 8 \times 7} \]

Calculating \( 3! = 6 \) and \( 2! = 2 \):

\[ \text{Probability} = \frac{6 \times 2}{10 \times 9 \times 8 \times 7} = \frac{12}{5040} \]

Simplifying \( \frac{12}{5040} \):

\[ \text{Probability} = \frac{1}{420} \]

### Final Answer:
The probability that the word "ANAND" occurs without getting split is \( \frac{1}{420} \).

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