A tea party is arranged for 16 people along two sides of a large table with 8 chair on each side. Four men want to sit on one particular side and two on the other side. The number of ways in which they can be seated is

Options

A. a. `""^(48)C_r`
B. b. `68!`
C. c. `(8!.8!10!)/(4!6!)`
D. d. none of these

Answer

To solve the problem of seating 16 people at a tea party with specific seating arrangements for men, we can break down the solution step by step.

### Step-by-Step Solution:

1. **Understanding the Arrangement**:
- There are 16 people in total.
- The seating is arranged on two sides of a table, with 8 chairs on each side.
- We need to seat 4 men on one side (let's call it Side A) and 2 men on the other side (Side B).

2. **Seating on Side A**:
- We need to choose 4 chairs out of the 8 available on Side A for the 4 men.
- The number of ways to choose 4 chairs from 8 is given by the combination formula \( C(n, r) = \frac{n!}{r!(n-r)!} \).
- Thus, the number of ways to choose 4 chairs from 8 is \( C(8, 4) \).

3. **Arranging the 4 Men on Side A**:
- Once the chairs are chosen, the 4 men can be arranged in those chairs.
- The number of ways to arrange 4 men in 4 chairs is \( 4! \).

4. **Seating on Side B**:
- For Side B, we need to choose 2 chairs from the remaining 8 chairs (since 4 chairs are already occupied on Side A).
- The number of ways to choose 2 chairs from 8 is \( C(8, 2) \).

5. **Arranging the 2 Men on Side B**:
- The 2 men can be arranged in the chosen chairs on Side B in \( 2! \) ways.

6. **Seating the Remaining People**:
- After seating the 6 men (4 on Side A and 2 on Side B), there are 10 people left.
- These 10 people can be seated in the remaining 10 chairs (4 on Side A and 6 on Side B) in \( 10! \) ways.

7. **Calculating the Total Arrangements**:
- The total number of arrangements can be calculated by multiplying the number of ways to choose and arrange the men on both sides and the arrangements of the remaining people:
\[ \text{Total Ways} = C(8, 4) \times 4! \times C(8, 2) \times 2! \times 10! \]

### Final Calculation:
Now, substituting the values:
- \( C(8, 4) = \frac{8!}{4!4!} = 70 \)
- \( 4! = 24 \)
- \( C(8, 2) = \frac{8!}{2!6!} = 28 \)
- \( 2! = 2 \)
- \( 10! = 3628800 \)

Putting it all together:
\[ \text{Total Ways} = 70 \times 24 \times 28 \times 2 \times 3628800 \]