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18 guests have to be seated, half on each side of a long table. 4 particular guests desire to sit on one particular side and 3 others on the other side. Number of ways in which the sitting arrangements can be made is:

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$\frac{18!}{3! 4!}$

$\frac{11! {(9!)}^{2}}{5! 6!}$

$\frac{2 \times 11! {(9!)}^{2}}{5! 6!}$

$\frac{11!}{5! 6!}$

answer is B.

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Detailed Solution

Since 4 particular guests have already chosen a particular side and three others on the other side, we have to choose 5 guests from remaining 11 to sit on the one side, which can be selected in C(11, 5) ways.

Remaining guests take their seat automatically on the other side hence does not come in calculation.

So total arrangement $= C (11, 5) \times 9! \times 9!$

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