We will say that a rearrangement of the letters of a word has no [i]fixed letters[/i] if, when the rearrangement is placed directly below the word, no column has the same letter repeated. For instance $HBRATA$ is a rearragnement with no fixed letter of $BHARAT$. How many distinguishable rearrangements with no fixed letters does $BHARAT$ have? (The two $A$s are considered identical.)
To solve the problem of finding the number of distinguishable rearrangements of the word "BHARAT" with no fixed letters, we need to consider the concept of derangements. A derangement is a permutation of a set where no element appears in its original position.

Given the word "BHARAT", we have the letters B, H, A, R, A, T. Note that the two A's are identical.

1. **Calculate the total number of derangements for 6 letters:**
   The formula for the number of derangements \( !n \) of \( n \) objects is given by:
   \[
   !n = n! \sum_{i=0}^{n} \frac{(-1)^i}{i!}
   \]
   For \( n = 6 \):
   \[
   !6 = 6! \left( \frac{(-1)^0}{0!} + \frac{(-1)^1}{1!} + \frac{(-1)^2}{2!} + \frac{(-1)^3}{3!} + \frac{(-1)^4}{4!} + \frac{(-1)^5}{5!} + \frac{(-1)^6}{6!} \right)
   \]
   \[
   !6 = 720 \left( 1 - 1 + \frac{1}{2} - \frac{1}{6} + \frac{1}{24} - \frac{1}{120} + \frac{1}{720} \right)
   \]
   \[
   !6 = 720 \left( 1 - 1 + 0.5 - 0.1667 + 0.0417 - 0.0083 + 0.0014 \right)
   \]
   \[
   !6 = 720 \left( 0.3681 \right) \approx 265
   \]

2. **Adjust for the identical letters:**
   Since the two A's are identical, we need to divide the total number of derangements by the number of permutations of the two A's:
   \[
   \text{Number of distinguishable derangements} = \frac{!6}{2!} = \frac{265}{2} = 132.5
   \]
   However, this result is not an integer, indicating a mistake in the initial calculation or approach.

3. **Recalculate using a combinatorial approach:**
   Let's use a combinatorial approach to count the derangements directly:
   - Place the two A's in any two of the six positions. There are \(\binom{6}{2} = 15\) ways to choose these positions.
   - For the remaining four positions (B, H, R, T), we need to find the derangements of these four letters.

4. **Calculate derangements for 4 letters:**
   Using the derangement formula for 4 letters:
   \[
   !4 = 4! \left( \frac{(-1)^0}{0!} + \frac{(-1)^1}{1!} + \frac{(-1)^2}{2!} + \frac{(-1)^3}{3!} + \frac{(-1)^4}{4!} \right)
   \]
   \[
   !4 = 24 \left( 1 - 1 + \frac{1}{2} - \frac{1}{6} + \frac{1}{24} \right)
   \]
   \[
   !4 = 24 \left( 1 - 1 + 0.5 - 0.1667 + 0.0417 \right)
   \]
   \[
   !4 = 24 \left( 0.375 \right) = 9
   \]

5. **Combine the results:**
   \[
   \text{Total number of derangements} = 15 \times 9 = 135
   \]

The final answer is \(\boxed{135}\).