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If 20 persons were invited for a party, in how many ways will two particular persons be seated on either side of the host in a circular arrangement?

According to me the answer should be $17!.2!$. But the given answer is $18!.2!$.

If we consider the guest and the host as one unit and let them take the first three chairs the other 17 can be occupied by 17! ways and the two particular persons can then rearrange them by 2! ways.

What am i doing wrong?

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when you treat the two guests and the host as one unit then you are left with $19$ entities to arrange ($18$ remaining guests $+$ this combined unit). The number of ways to arrange them is $18!$ and then you can permute the two special guests in $2!$ ways. In all $18! 2!$.

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  • $\begingroup$ If I am left with 18 entitities should not i be making (n-1)! =(18-1)! arrangements? This is circular permutation problem $\endgroup$ – archangel89 May 4 '15 at 7:23
  • $\begingroup$ @archangel89 You are right about circular arrangement and I had factored that but unfortunately I typed it incorrectly. Note there are $20$ guests and $1$ host so in all $21$ people. Now see the solution (I have edited it) $\endgroup$ – Anurag A May 4 '15 at 7:28
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    $\begingroup$ Oh thank you. I forgot host is different from the guests. $\endgroup$ – archangel89 May 4 '15 at 7:29

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