Question

A gentleman invites a party of $m+n(m\ne n)$ friends to a dinner & places $m$ at one table and $n$ at another, the table being round. If the clockwise & anticlockwise arrangements are not to be distinguished and assuming sufficient space on both tables, then the number of ways in which he can arrange the guest is

• A. $\frac{(m+n)!}{4mn}$
• B. $\frac{1}{2}\frac{(m+n)!}{4mn}$
• C. $2\frac{(m+n)!}{4mn}$
• D. none

Answer 1

Answer: (D)

none

Gentleman have total $(m+n)$ friends. We can choose $m$ people by ${}^{m+n}C=\frac{(m+n)!}{m!\times n!}$ ways. Remaining $n$ friends will sit on other table.

We can arrange $m$ people by $(m-1)!$ and $n$ people by $(n-1)!$

Combining all we get $\frac{(m+n)!}{m!\times n!}\times (m-1)!\times (n-1)!$ = $\frac{(m+n)!}{m\times n}$