Step 1 : Understanding the Question:
This problem is in the area of combinatorics, involving both grouping (combinations) and circular permutations. We need to determine the total number of ways to distribute 10 friends into two groups of 4 and 6, and then arrange them around two distinct circular tables.
Step 2 : Key Formulas and Approach:
First, we select which friends sit at which table using the combination formula:
\[ \binom{n}{r} = \frac{n!}{r!(n-r)!} \]
Second, we use the circular permutation formula. The number of unique ways to arrange \( m \) distinct items around a circular table is given by:
\[ P_{\text{circle}} = (m-1)! \]
Finally, we apply the fundamental counting principle by multiplying the number of ways to group the friends by the number of ways to arrange them at each table.
Step 3 : Detailed Explanation:
Step 1: Select the 4 friends who will sit at the first table. Out of 10 friends, the number of ways to choose 4 is:
\[ \binom{10}{4} = \frac{10!}{4! \times 6!} \]
The remaining 6 friends are automatically assigned to the second table in \( \binom{6}{6} = 1 \) way.
Step 2: Calculate the circular arrangements at each table.
The 4 friends at the first round table can be arranged in \( (4-1)! = 3! = 6 \) ways.
The 6 friends at the second round table can be arranged in \( (6-1)! = 5! = 120 \) ways.
Step 3: Multiply the selection ways and the arrangement ways to find the total:
\[ \text{Total} = \binom{10}{4} \times 3! \times 5! \]
\[ \text{Total} = \frac{10!}{4! \times 6!} \times 3! \times 5! \]
Step 4: Simplify the factorial expression:
\[ \text{Total} = \frac{10!}{(4 \times 3!) \times (6 \times 5!)} \times 3! \times 5! \]
\[ \text{Total} = \frac{10!}{24 \times 6 \times 5!} \times 6 \times 5! = \frac{10!}{24} \]
Step 4 : Final Answer:
The total number of ways to arrange the guests is \( \frac{10!}{24} \), which corresponds to option (B).