Question

A gentleman invites a party of $m+n(m\ne n)$ friends to a dinner and places $m$ at one table $T_1$ and $n$ at another table $T_2$, the table being around. If not all people shall have the same neighbour in any two arrangement, then the number of ways in which he can arrange the guest, is

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

Answer 1

Answer: (A), (D)

The correct options are
A $\frac{(m+n)!}{4mn}$
D $None$

There are $(m+n)$ friends out of which $m$ sit at one table and $n$ at one table $={}^{m+n}{C}_m,$ rest can be selected in $1$ way.

$m$ can be rearranged in $(m-1)!$ ways $and$ $n$ people in $(n-1)!$

Also given in each arrangement not all will have same neighbour.

Clockwise $and$ anticlockwise is considered same.

$\Rightarrow$ Total $={}^{m+n}{C}_m\times \frac{(m-1)!}{2}\times \frac{(n-1)!}{2}=\frac{(m+n)!}{4mn}$

Hence, the answer is $\frac{(m+n)!}{4mn}.$