A man invites a party to (m+n) friends to dinner and places m at one round table and n at another. The number of ways of arranging the guests is

Options

A. `((m+n)!)/(m!n!)`
B. `((m+n)!)/((m-1)!(n-1)!)`
C. `(m-1)!(n-1)!`
D. none of these

Answer

To solve the problem of arranging (m+n) friends at two round tables, one with m friends and the other with n friends, we can follow these steps:

### Step-by-Step Solution:

1. **Select m Friends for the First Table:**
We need to choose m friends from a total of (m+n) friends. The number of ways to do this is given by the combination formula:
\[ \binom{m+n}{m} \]

2. **Select n Friends for the Second Table:**
After selecting m friends, the remaining friends will automatically be n (since m + n - m = n). The number of ways to select these n friends from the remaining n friends is:
\[ \binom{n}{n} = 1 \]
(There is only one way to select all n friends.)

3. **Arrange m Friends at the First Round Table:**
For arranging m friends at a round table, we fix one friend and arrange the remaining (m-1) friends. The number of arrangements is:
\[ (m-1)! \]

4. **Arrange n Friends at the Second Round Table:**
Similarly, for arranging n friends at the second round table, we fix one friend and arrange the remaining (n-1) friends. The number of arrangements is:
\[ (n-1)! \]

5. **Combine All the Steps:**
Now, we combine all the ways to select and arrange the friends:
\[ \text{Total Ways} = \binom{m+n}{m} \times (m-1)! \times 1 \times (n-1)! \]

6. **Simplifying the Expression:**
We can simplify this expression. The combination can be expressed as:
\[ \binom{m+n}{m} = \frac{(m+n)!}{m!n!} \]
Thus, the total number of arrangements becomes:
\[ \text{Total Ways} = \frac{(m+n)!}{m!n!} \times (m-1)! \times (n-1)! \]

7. **Final Simplification:**
Now, we can cancel out the factorials:
\[ = \frac{(m+n)!}{m \cdot n} \]

### Final Answer:
The total number of ways to arrange the guests is:
\[ \frac{(m+n)!}{m \cdot n} \]