7 sisters dine at a round table. They dine together till each of them dine with different neighbours i.e., they do not like to done dine with same neighbours in any two arrangements. In a year how many days they dine together?

Options

A. 180
B. 504
C. 720
D. 360

Answer

To solve the problem of how many days the 7 sisters can dine together at a round table without repeating any arrangement of neighbors, we can follow these steps:

### Step-by-Step Solution:

1. **Understanding the Circular Arrangement**:
When arranging people in a circle, the formula for the number of arrangements is given by (n-1)!, where n is the number of people. This is because one person's position can be fixed to eliminate the effect of rotations.

Here, we have 7 sisters, so:
\[ \text{Number of arrangements} = (7-1)! = 6! \]

2. **Calculating 6!**:
Now we calculate 6!:
\[ 6! = 6 \times 5 \times 4 \times 3 \times 2 \times 1 = 720 \]

3. **Considering Neighbor Restrictions**:
Since the sisters do not want to repeat any arrangement with the same neighbors, we need to account for the fact that each arrangement can be flipped (i.e., if A is next to B, the arrangement can also be B next to A). Thus, each arrangement has a duplicate.

Therefore, we divide the total arrangements by 2:
\[ \text{Unique arrangements} = \frac{6!}{2} = \frac{720}{2} = 360 \]

4. **Calculating the Number of Days**:
Since there are 360 unique arrangements, and assuming they dine together every day, the number of days they can dine together in a year (365 days) is limited by the number of unique arrangements. Thus, they can dine together for:
\[ \text{Days they can dine together} = 360 \text{ days} \]

### Final Answer:
The number of days the 7 sisters can dine together without repeating any arrangement is **360 days**.

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