Question

A person invites 10 friends to dinner and places them such that 4 are at one round table and 6 are at another round table. The total number of ways in which he can arrange the guests is:

• A. \( 10!/6! \)
• B. \( 10!/24 \)
• C. \( 9!/24 \)
• D. \( \text{None of these} \)

Hint:

Always remember to partition (choose) the groups first before applying circular permutation formulas to each subset.

Answer 1

The Correct Option is B

Concept: The number of ways to arrange $n$ distinct objects in a circle is $(n-1)!$. When we need to partition a group into smaller sets and arrange them at separate tables, we must consider both the selection of guests for each table and the circular arrangement at each table.

Step 1: Select the guests for each table. We need to choose 4 friends out of 10 to sit at the first table, and the remaining 6 will sit at the second table. The number of ways to do this is: \[ \binom{10}{4} = \frac{10!}{4! \times 6!} \]

Step 2: Arrange the guests at the round tables. 1. The 4 friends at the first table can be arranged in $(4-1)! = 3! = 6$ ways. 2. The 6 friends at the second table can be arranged in $(6-1)! = 5! = 120$ ways.

Step 3: Calculate the total number of ways. Total ways = (Ways to select) $\times$ (Ways to arrange at Table 1) $\times$ (Ways to arrange at Table 2) \[ \text{Total} = \binom{10}{4} \times 3! \times 5! \] \[ \text{Total} = \frac{10!}{4! \times 6!} \times 3! \times 5! \] \[ \text{Total} = \frac{10!}{(4 \times 3!) \times 6} \times 3! \times 120 = \frac{10!}{4 \times 6 \times 6} \times 6 \times 120 = \frac{10!}{24} \times \frac{120}{6} = \frac{10!}{24} \times 20 \] Actually, simplifying directly: \[ \frac{10!}{4! \times 6!} \times 3! \times 5! = \frac{10!}{ (4 \times 3!) \times 6} \times 3! \times 5! = \frac{10! \times 3! \times 5!}{24 \times 6 \times 5!} = \frac{10! \times 6}{24 \times 6} = \frac{10!}{24} \] Since $10!/24$ is mathematically correct and appears to be Option (2), the answer is (2).