Q59 / 100 · BITSAT 2024 (Memory Based Paper 3) · Mathematics

A person invites a party of 10 friends at dinner and place 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
    $\frac{10!}{6!}$
  • Correct answer
    B
    $\frac{10!}{24}$
  • C
    $\frac{9!}{24}$
  • D
    None of these

Solution

Total person $=10$ Two groups are formed containing 4 and 6 persons. $\therefore$ Total ways of formation of groups $=\frac{10!}{4!6!}$ Now, 4 persons can be arranged on round table $=(4-1)!=3!$ and 6 persons can be arranged on round table $=(6-1)$ ! $=5!$ $\therefore$ Total ways of arranging the guest $=\frac{10!}{4!6!} \times 3!\times 5!=\frac{10!}{24}$

Hence, the correct answer is option: 'B'.

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