A gentlemen hosts a party of (m+n) guests and places 'm' at one round table and the remaining 'n' at the other round table. Number of ways the guests can be arranged is

Options

A. `((m +n)!)/(m*n)`
B. `((m +n))/(m*n)`
C. `((m +n))/(m!*n!)`
D. `((m +n))/(m^(n))`

Answer

To find the number of ways to arrange (m+n) guests at two round tables, where 'm' guests are seated at one table and 'n' guests at another, we can follow these steps:

### Step-by-Step Solution:

1. **Total Guests**: The total number of guests is \( m + n \).

2. **Choosing Guests for the First Table**: We need to choose 'm' guests from the total of \( m + n \) guests to sit at the first round table. The number of ways to choose 'm' guests from \( m+n \) is given by the combination formula:
\[ \binom{m+n}{m} = \frac{(m+n)!}{m! \cdot n!} \]

3. **Arranging Guests at the First Round Table**: Once we have chosen 'm' guests for the first table, they can be arranged in a circular manner. The number of ways to arrange 'm' guests in a round table is given by:
\[ (m-1)! \]
(This is because in circular arrangements, one position is fixed to avoid counting rotations as different arrangements.)

4. **Arranging Guests at the Second Round Table**: The remaining 'n' guests will sit at the second table. The number of ways to arrange 'n' guests in a circular manner is:
\[ (n-1)! \]

5. **Total Arrangements**: The total number of arrangements of guests at both tables is the product of the number of ways to choose the guests and the arrangements at each table:
\[ \text{Total Arrangements} = \binom{m+n}{m} \cdot (m-1)! \cdot (n-1)! \]

6. **Substituting the Combination**: Now substituting the combination formula:
\[ \text{Total Arrangements} = \frac{(m+n)!}{m! \cdot n!} \cdot (m-1)! \cdot (n-1)! \]

7. **Simplifying the Expression**: We can simplify the expression:
\[ = \frac{(m+n)!}{m \cdot n} \quad \text{(since } m! = m \cdot (m-1)! \text{ and } n! = n \cdot (n-1)!) \]

Thus, the final number of ways the guests can be arranged is:
\[ \frac{(m+n)!}{m \cdot n} \]

### Final Answer:
The number of ways the guests can be arranged is \( \frac{(m+n)!}{m \cdot n} \).