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A man invites 10 friends for a dinner party. In how many they can be seated around a circular table so that two particular friends on either side of man?

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10!

$8! \times 2!$

answer is D.

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Detailed Solution

Let M be the man and $F_{1}, F_{2}$ be two particular friends

$F_{1} M F_{2}$ considered to be a single person

We have 9 in all

They can be arranged in 8! and $F_{1}, F_{2}$ can be interchange in 2 ways

Total $= 8! \times 2$ .

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