In how many ways can 8 persons be seated at a round table so that all shall not have the same neighbours in any two arrangements?

Answer

To solve the problem of seating 8 persons at a round table such that no two arrangements have the same neighbors, we can follow these steps:

### Step-by-Step Solution:

1. **Fix One Person**: In circular permutations, we can fix one person to eliminate the effect of rotations. Let's fix one person at the table. This leaves us with 7 persons to arrange.

2. **Arrange Remaining Persons**: The remaining 7 persons can be arranged in a linear fashion around the fixed person. The number of ways to arrange 7 persons is given by \(7!\) (7 factorial).

\[ 7! = 5040 \]

3. **Consider Neighbor Restrictions**: The problem states that we want to ensure that no two arrangements have the same neighbors. This means that if we have a specific arrangement, we cannot have another arrangement that has the same neighbors for any person.

4. **Adjust for Neighboring Conditions**: Since we have fixed one person, let's say person A. The neighbors of A in any arrangement will be the two persons directly next to A. If we consider the arrangement of the remaining 7 persons, we realize that for each arrangement, the neighbors of A will be the same.

5. **Divide by Neighbor Permutations**: Since we want to ensure that we do not count arrangements that are simply rotations of each other, we need to divide the total arrangements by the number of ways to arrange the neighbors of A. There are 2 ways to arrange the neighbors of A (one on the left and one on the right). Thus, we divide the total arrangements by 2.

\[ \text{Total arrangements} = \frac{7!}{2} = \frac{5040}{2} = 2520 \]

### Final Answer:
The total number of ways to arrange 8 persons at a round table such that no two arrangements have the same neighbors is **2520**.