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A gentleman invites a party of $m + n (m \neq n)$ friends to a dinner and place m at one table $T_{1} a n d n$ at another table $T_{2},$ both tables being round. If not all people shall have the same neighbor in any two arrangement, then the number of ways in which he can arrange the guests, is:

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$2 \frac{(m + n)!}{m n}$

$\frac{1}{2} \frac{(m + n)!}{m n}$

none of these

$\frac{(m + n)!}{4 m n}$

answer is A.

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Detailed Solution

$\frac{(m + n)!}{m! n!} . \frac{(m - 1)!}{2} . \frac{(n - 1)!}{2}$

$\Rightarrow a$

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