Question

A person invites a party of $10$ friends at dinner and place them so that $4$ are on one round table and $6$ on the other round table. The 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

Answer 1

Answer: (B)

Selection of $6$ guests $={ }^{10} C_{6}$
Permutation of $6$ on round table $=5 !$
Permutation of $4$ on round table $=3$
Then, total number of arrangements
$={ }^{10} C_{6} \cdot 5 ! \cdot 3 !=\frac{(10) !}{6 ! \cdot 4 !} \cdot 5 ! \cdot 3 !$
$=\frac{(10) !}{24}$