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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
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The correct answer is:
$\frac{(10) !}{24}$
Selection of 6 guests $=10 \mathrm{C} 6$
Permutation of 6 on round table $=5 !$
Permutation of 4 on round table $=3 !$
Then, total number of arrangements $=10 \mathrm{C} 6.5 ! .3 !$
$=\frac{(10) !}{6 ! 4 !} \cdot 5 ! .3 !=\frac{(10) !}{24}$
Permutation of 6 on round table $=5 !$
Permutation of 4 on round table $=3 !$
Then, total number of arrangements $=10 \mathrm{C} 6.5 ! .3 !$
$=\frac{(10) !}{6 ! 4 !} \cdot 5 ! .3 !=\frac{(10) !}{24}$
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