In "BHARAT" how many of these B and H are never together?

Answer

To solve the problem of how many arrangements of the letters in "BHARAT" have the letters B and H never together, we can follow these steps:

### Step 1: Calculate the total arrangements of the letters in "BHARAT".

The word "BHARAT" consists of 6 letters where 'A' is repeated twice. The formula for the total arrangements of letters when there are repetitions is given by:

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

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

For "BHARAT":
- Total letters, \( n = 6 \)
- The letter 'A' is repeated 2 times.

Thus, the total arrangements are:

\[ \text{Total arrangements} = \frac{6!}{2!} = \frac{720}{2} = 360 \]

### Step 2: Calculate the arrangements where B and H are together.

To find the arrangements where B and H are together, we can treat B and H as a single entity or block. This block can be denoted as (BH) or (HB).

Now, we have the following entities to arrange:
- (BH) or (HB)
- A
- A
- R
- T

This gives us a total of 5 entities: (BH), A, A, R, T.

The arrangements of these 5 entities, considering the repetition of 'A', is given by:

\[ \text{Arrangements with B and H together} = \frac{5!}{2!} = \frac{120}{2} = 60 \]

### Step 3: Calculate the arrangements where B and H are NOT together.

To find the arrangements where B and H are not together, we subtract the arrangements where they are together from the total arrangements:

\[ \text{Arrangements where B and H are NOT together} = \text{Total arrangements} - \text{Arrangements with B and H together} \]

Substituting the values we calculated:

\[ \text{Arrangements where B and H are NOT together} = 360 - 60 = 300 \]

### Final Answer:

The number of arrangements of the letters in "BHARAT" where B and H are never together is **300**.

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