A tea party is arranged for 16 persons along two sides of a long table with 8 chairs on each side. Four persons wish to sit on one particular and two on the other side. In how many ways can they be seated?

Explanation

To solve the problem of seating 16 persons at a tea party along two sides of a long table with specific seating preferences, we can break down the solution into clear steps.

### Step-by-Step Solution:

1. **Identify the seating arrangement**: We have 16 persons and 16 chairs (8 on each side of the table). Among these, 4 persons want to sit on one side (let's call it Side A) and 2 persons on the other side (Side B).

2. **Select chairs for the 4 persons on Side A**:
- We need to choose 4 chairs from the 8 available on Side A. The number of ways to choose 4 chairs from 8 is given by the combination formula \( \binom{n}{r} \), which represents the number of ways to choose \( r \) items from \( n \) items without regard to the order of selection.
- Therefore, the number of ways to choose 4 chairs from 8 is \( \binom{8}{4} \).

3. **Arrange the 4 persons in the selected chairs**:
- Once we have chosen the 4 chairs, we can arrange the 4 persons in these chairs. The number of ways to arrange 4 persons is given by \( 4! \) (4 factorial).
- Thus, the total arrangements for the 4 persons on Side A is \( \binom{8}{4} \times 4! \).

4. **Select chairs for the 2 persons on Side B**:
- Next, we need to choose 2 chairs from the 8 available on Side B. The number of ways to choose 2 chairs from 8 is \( \binom{8}{2} \).

5. **Arrange the 2 persons in the selected chairs**:
- After selecting the chairs, we can arrange the 2 persons in these chairs. The number of ways to arrange 2 persons is \( 2! \).
- Therefore, the total arrangements for the 2 persons on Side B is \( \binom{8}{2} \times 2! \).

6. **Arrange the remaining 10 persons**:
- After seating the 6 persons (4 on Side A and 2 on Side B), there are 10 persons left to be seated in the remaining 10 chairs.
- The number of ways to arrange these 10 persons in 10 chairs is \( 10! \).

7. **Combine all the arrangements**:
- The total number of ways to arrange all persons is the product of the arrangements calculated in the previous steps:
\[ \text{Total Ways} = \left( \binom{8}{4} \times 4! \right) \times \left( \binom{8}{2} \times 2! \right) \times 10! \]

### Final Calculation:
- Using the known values:
- \( \binom{8}{4} = 70 \)
- \( 4! = 24 \)
- \( \binom{8}{2} = 28 \)
- \( 2! = 2 \)
- \( 10! = 3628800 \)

- Plugging these values into the equation:
\[ \text{Total Ways} = (70 \times 24) \times (28 \times 2) \times 3628800 \]

- Calculating:
\[ \text{Total Ways} = 1680 \times 56 \times 3628800 \]

- Finally, compute the total:
\[ \text{Total Ways} = 1680 \times 56 = 94080 \]
\[ \text{Total Ways} = 94080 \times 3628800 = 341373696000 \]