Eighteen guests are to be seated half on each side of a long table. Four particular guests desire to sit on particular side and three others on other side of the table. The number of ways in which the seating arrangements can be made is

Options

A. `((11!))/((5!)(6!))((9!))^(2)`
B. `((11!))/((5!)(6!))((9!))^(2)*2!`
C. `((11!))/((5!)(6!))(2!)`
D. `((11!))/((5!)(6!))((9!))*(2!)`

Answer

To solve the problem of seating 18 guests with specific seating preferences, we can break down the solution into clear steps:

### Step-by-Step Solution:

1. **Understanding the Problem**:
- We have 18 guests in total.
- 9 guests will sit on Side A and 9 guests will sit on Side B of the table.
- 4 specific guests want to sit on Side A.
- 3 specific guests want to sit on Side B.

2. **Determine Remaining Guests**:
- After seating the 4 guests on Side A and the 3 guests on Side B, we have:
- Remaining guests = 18 - 4 - 3 = 11 guests.

3. **Selecting Guests for Each Side**:
- We need to select additional guests to fill the sides:
- Side A needs 5 more guests (since 4 are already seated).
- Side B needs 6 more guests (since 3 are already seated).
- We can choose 5 guests from the remaining 11 guests to sit on Side A.

4. **Calculating Combinations**:
- The number of ways to choose 5 guests from 11 is given by the combination formula:
\[ \binom{11}{5} = \frac{11!}{5!(11-5)!} = \frac{11!}{5!6!} \]

5. **Seating Arrangements**:
- Once we have chosen 9 guests for Side A (4 specific + 5 chosen) and 9 guests for Side B (3 specific + 6 chosen), we can arrange them.
- The number of ways to arrange 9 guests on Side A is \(9!\) and the number of ways to arrange 9 guests on Side B is also \(9!\).

6. **Total Arrangements**:
- The total number of seating arrangements is given by:
\[ \text{Total arrangements} = \binom{11}{5} \times 9! \times 9! \]

7. **Final Calculation**:
- Substitute the values into the formula:
\[ \text{Total arrangements} = \frac{11!}{5!6!} \times 9! \times 9! \]

### Final Answer:
Thus, the total number of ways in which the seating arrangements can be made is:
\[ \frac{11!}{5!6!} \times (9!)^2 \]