Harsha invites 13 guests to a dinner and places 8 of them at one table and remaining 5 at the other, the table being round. The number of ways he can arrange the guests is

Options

A. `(11!)/(40)`
B. `9!`
C. `(13!)/(40)`
D. `(12!)/(40)`

Answer

To solve the problem of arranging 13 guests at two round tables, we will follow these steps:

### Step 1: Choose Guests for Each Table
We need to select 8 guests out of the 13 to sit at the first table. The number of ways to choose 8 guests from 13 can be calculated using the combination formula:

\[ \text{Number of ways to choose 8 guests} = \binom{13}{8} = \frac{13!}{8! \cdot 5!} \]

### Step 2: Arrange Guests at the Round Tables
Since the tables are round, we need to arrange the guests in a circular manner. For a round table, the number of arrangements of \( n \) guests is given by \( (n-1)! \).

- For the table with 8 guests, the number of arrangements is:
\[ (8-1)! = 7! \]

- For the table with 5 guests, the number of arrangements is:
\[ (5-1)! = 4! \]

### Step 3: Combine the Results
Now, we need to combine the results from the two steps. The total number of arrangements can be calculated by multiplying the number of ways to choose the guests by the number of arrangements at each table:

\[ \text{Total arrangements} = \binom{13}{8} \cdot (7!) \cdot (4!) \]

Substituting the combination formula:

\[ \text{Total arrangements} = \frac{13!}{8! \cdot 5!} \cdot 7! \cdot 4! \]

### Step 4: Simplify the Expression
We can simplify the expression as follows:

\[ = \frac{13!}{8! \cdot 5!} \cdot 7! \cdot 4! = \frac{13!}{8!} \cdot \frac{7!}{5!} \cdot 4! \]

Since \( \frac{7!}{5!} = 7 \cdot 6 \) and \( \frac{13!}{8!} = 13 \cdot 12 \cdot 11 \cdot 10 \cdot 9 \), we can further simplify:

\[ = 13 \cdot 12 \cdot 11 \cdot 10 \cdot 9 \cdot 7 \cdot 6 \cdot 4! \]

### Step 5: Calculate the Final Answer
The final answer can be calculated by evaluating the above expression.

### Final Result
The total number of ways Harsha can arrange the guests is:

\[ \text{Total arrangements} = \frac{13!}{8! \cdot 5!} \cdot 7! \cdot 4! = 13 \cdot 12 \cdot 11 \cdot 10 \cdot 9 \cdot 7 \cdot 6 \cdot 24 \]