Question

A person invites a party of 10 friends at dinner and places so that 4 are on one round table and 6 on the other round table. Total number of ways in which he can arrange the guests is:

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

Hint:

In problems involving round table arrangements, subtract 1 from the number of people to account for rotational symmetry.

Answer 1

The Correct Option is B

Step 1: {Total number of people} 
The total number of people is \( 10 \), and they are divided into two groups: one group of 4 persons and another group of 6 persons. 
Step 2: {Number of ways to form groups} 
The number of ways to form the two groups is \( \frac{10!}{4!6!} \). 
Step 3: {Arrangements on the round tables} 
For the group of 4 persons, the number of arrangements on a round table is \( (4-1)! = 3! = 6 \). For the group of 6 persons, the number of arrangements on a round table is \( (6-1)! = 5! = 120 \). 
Step 4: {Total ways to arrange the guests} 
The total number of ways to arrange the guests is: \[ \frac{10!}{4!6!} \times 6 \times 120 = \frac{10!}{24}. \] Thus, the total number of ways is \( \frac{10!}{24} \), which matches option (B).