The number of ways in which four persons be seated at a round table, so that all shall not have the same neighbours in any two arrangements,is

Options

A. 24
B. 6
C. 3
D. 4

Answer

To solve the problem of seating four persons at a round table such that no two arrangements have the same neighbors, we can follow these steps:

### Step-by-Step Solution:

1. **Understanding Circular Arrangements**:
In a circular arrangement, one person can be fixed to eliminate the effect of rotations. This means that for 'n' persons, the number of arrangements is given by (n - 1)!.

2. **Fixing One Person**:
Here, we have 4 persons (let's call them A, B, C, and D). We can fix one person (say A) in one position.

3. **Calculating Arrangements**:
The remaining three persons (B, C, and D) can be arranged in the remaining seats. The number of arrangements for these three persons is (4 - 1)! = 3! = 6.

4. **Considering Neighbor Conditions**:
The problem states that no two arrangements should have the same neighbors. This means that if we have one arrangement, we should not consider its reverse arrangement as valid.

5. **Dividing by Two**:
Since each arrangement can be mirrored (flipped), we need to divide the total arrangements by 2 to ensure that we only count one of the two possible arrangements (clockwise or anti-clockwise).

Therefore, we take the total arrangements (6) and divide by 2:
\[ \text{Valid arrangements} = \frac{6}{2} = 3. \]

6. **Final Answer**:
Thus, the number of ways in which four persons can be seated at a round table, ensuring that no two arrangements have the same neighbors, is 3.