Question:medium

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:

Show Hint

In problems involving round table arrangements, subtract 1 from the number of people to account for rotational symmetry.
Updated On: Jan 13, 2026
  • \( \frac{10!}{6!} \)
  • \( \frac{10!}{24} \)
  • \( \frac{9!}{24} \)
  • None of these
Show Solution

The Correct Option is B

Solution and Explanation

Step 1: {Total number of people}
There are \( 10 \) individuals in total, partitioned into two subgroups of \( 4 \) and \( 6 \) members respectively.
Step 2: {Number of ways to form groups}
The number of distinct combinations for forming these two groups is calculated as \( \frac{10!}{4!6!} \).
Step 3: {Arrangements on the round tables}
For the subgroup of \( 4 \) individuals, circular permutations on a round table yield \( (4-1)! = 3! = 6 \) possible arrangements. For the subgroup of \( 6 \) individuals, circular permutations on a round table yield \( (6-1)! = 5! = 120 \) possible arrangements.
Step 4: {Total ways to arrange the guests}
The total number of arrangements for all guests is the product of group formation and table arrangements: \[ \frac{10!}{4!6!} \times 6 \times 120 = \frac{10!}{24}. \] Consequently, the total number of ways to arrange the guests is \( \frac{10!}{24} \), aligning with option (B).

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