A person invites a party of 10 friends at dinner and place so that...

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 $=\frac{10!}{4!6!} \times 3!\times 5!=\frac{10!}{24}$