There are eight rooms on the first floor of a hotel, with four rooms on each side of the corridor, symmetrical situated (that is each room is exactly opposite to one other room). Four guests have to be accommodation In four of the eight rooms (that is, one in each) such that no two guests are in adjacent rooms or in opposite rooms. If N is the number of ways in which guests can be accommodated. Then the value of `N/6` is

Answer

To solve the problem, we need to determine how many ways 4 guests can be accommodated in 4 out of 8 rooms in a hotel, ensuring that no two guests are in adjacent or opposite rooms.

### Step-by-Step Solution:

1. **Understanding the Room Arrangement**:
- We have 8 rooms arranged as follows:
- Side A: Room 1, Room 2, Room 3, Room 4
- Side B: Room 5, Room 6, Room 7, Room 8
- Rooms 1 and 5 are opposite, 2 and 6 are opposite, and so on.

2. **Identifying Valid Room Combinations**:
- Since no two guests can be in adjacent rooms or in opposite rooms, we need to select rooms in such a way that they are separated.
- We can visualize the arrangement as two groups:
- Group 1 (Side A): Rooms 1, 2, 3, 4
- Group 2 (Side B): Rooms 5, 6, 7, 8

3. **Choosing Rooms**:
- We can choose rooms from either Group 1 or Group 2, but not both since they are opposite.
- Let's denote the rooms chosen from Group 1 as A and from Group 2 as B.

4. **Case Analysis**:
- **Case 1**: Choose 2 rooms from Group 1 (A) and 2 rooms from Group 2 (B).
- **Case 2**: Choose 1 room from Group 1 (A) and 3 rooms from Group 2 (B).
- **Case 3**: Choose 3 rooms from Group 1 (A) and 1 room from Group 2 (B).

5. **Calculating Combinations**:
- **Case 1**: Choosing 2 rooms from Group 1 and 2 from Group 2:
- The valid combinations for Group 1 (A) are (1, 3), (1, 4), (2, 4) - which gives us 3 ways.
- The same applies for Group 2 (B) - also 3 ways.
- Total for this case = 3 (A) * 3 (B) = 9 ways.

- **Case 2**: Choosing 1 from Group 1 and 3 from Group 2:
- The valid combinations for Group 1 (A) are 1 way (only Room 1).
- The valid combinations for Group 2 (B) are (5, 7), (5, 8), (6, 8) - which gives us 3 ways.
- Total for this case = 1 (A) * 3 (B) = 3 ways.

- **Case 3**: Choosing 3 from Group 1 and 1 from Group 2:
- The valid combinations for Group 1 (A) are (1, 2, 4), (1, 3, 4), (2, 3, 4) - which gives us 3 ways.
- The valid combinations for Group 2 (B) are 1 way (only Room 5).
- Total for this case = 3 (A) * 1 (B) = 3 ways.

6. **Total Combinations**:
- Adding all the cases together:
- Case 1: 9 ways
- Case 2: 3 ways
- Case 3: 3 ways
- Total = 9 + 3 + 3 = 15 ways.

7. **Arranging Guests**:
- For each arrangement of rooms, the guests can be arranged in those rooms in 4! (factorial) ways.
- Therefore, total arrangements = 15 * 4! = 15 * 24 = 360.

8. **Finding N/6**:
- We need to find N/6, where N = 360.
- N/6 = 360 / 6 = 60.

### Final Answer:
The value of N/6 is **60**.