To solve the problem of how many arrangements of the word "BHARAT" have the letters B and H never together, we can follow these steps:
### Step 1: Calculate the Total Arrangements of the Word "BHARAT"
The word "BHARAT" consists of 6 letters where the letter 'A' is repeated twice. The total number of arrangements can be calculated using the formula for permutations of multiset:
\[
\text{Total arrangements} = \frac{n!}{p_1! \cdot p_2! \cdots p_k!}
\]
Where:
- \( n \) is the total number of letters,
- \( p_1, p_2, \ldots, p_k \) are the frequencies of the repeated letters.
For "BHARAT":
- Total letters \( n = 6 \)
- Repeated letters: A appears 2 times.
Thus, the total arrangements are:
\[
\text{Total arrangements} = \frac{6!}{2!} = \frac{720}{2} = 360
\]
### Step 2: Calculate the Arrangements with B and H Together
Next, we consider B and H as a single unit or block. If we treat "BH" as one letter, we then have the following letters to arrange: {BH, A, R, A, T}.
This gives us a total of 5 units (BH, A, R, A, T), where A is still repeated twice. The number of arrangements of these 5 units is:
\[
\text{Arrangements with BH together} = \frac{5!}{2!} = \frac{120}{2} = 60
\]
### Step 3: Account for Internal Arrangements of B and H
Since B and H can also be arranged among themselves (as "BH" or "HB"), we need to multiply the arrangements by 2:
\[
\text{Total arrangements with B and H together} = 60 \times 2 = 120
\]
### Step 4: Calculate the Arrangements with B and H Never Together
Finally, to find the arrangements where B and H are never together, we subtract the arrangements where they are together from the total arrangements:
\[
\text{Arrangements with B and H never together} = \text{Total arrangements} - \text{Arrangements with B and H together}
\]
\[
\text{Arrangements with B and H never together} = 360 - 120 = 240
\]
### Final Answer
Thus, the number of arrangements of the word "BHARAT" where B and H are never together is **240**.
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