Eighteen guests have to be seated half on each side of a long table. Four particular guests desire to sit on one particular side and three others on other side. Determine the number of ways in which the sitting, arrangements can be made.

Options

A. a. `18!`
B. b. `462 xx (9!)^2`
C. c. `(9!)^2`
D. d. `7!""xx (9!)^2`

Answer

To solve the problem of seating arrangements for 18 guests at a long table, where specific guests have preferences for which side they want to sit on, we can follow these steps:

### Step-by-Step Solution:

1. **Identify the Guests and Their Preferences**:
- Total guests = 18
- Guests on Side A (4 particular guests) = 4
- Guests on Side B (3 particular guests) = 3
- Remaining guests = 18 - 4 - 3 = 11 guests

2. **Determine the Number of Guests on Each Side**:
- Each side of the table will have 9 guests (half of 18).
- Side A will have the 4 particular guests and 5 more guests from the remaining 11.
- Side B will have the 3 particular guests and the remaining guests.

3. **Select Additional Guests for Each Side**:
- We need to select 5 guests from the remaining 11 guests to sit on Side A.
- The number of ways to choose 5 guests from 11 is given by the combination formula \( \binom{n}{r} \), where \( n \) is the total number of items to choose from, and \( r \) is the number of items to choose.

\[ \text{Ways to choose 5 guests from 11} = \binom{11}{5} = \frac{11!}{5! \cdot 6!} \]

4. **Calculate the Combinations**:
- Calculate \( \binom{11}{5} \):
\[ \binom{11}{5} = \frac{11 \times 10 \times 9 \times 8 \times 7}{5 \times 4 \times 3 \times 2 \times 1} = 462 \]

5. **Arrange Guests on Each Side**:
- After selecting the guests, we have 9 guests on each side (4 + 5 on Side A and 3 + 6 on Side B).
- The number of ways to arrange 9 guests on Side A is \( 9! \).
- The number of ways to arrange 9 guests on Side B is also \( 9! \).

6. **Total Arrangements**:
- The total number of arrangements is the product of the combinations and the arrangements on each side:
\[ \text{Total arrangements} = \binom{11}{5} \times 9! \times 9! = 462 \times 9! \times 9! \]

### Final Answer:
The total number of ways in which the seating arrangements can be made is \( 462 \times (9!)^2 \).