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At a wedding ceremony there are 12 people, including Ann, Ben, Clair, are to be seated in a ROUND table with 12 chairs. In how many ways can they sit if neither Ann nor Ben can sit next to Clair?

First I calculated the possibilities which all 3 can sit together(So I can deduct it later) i.e. 9!3! But how do I calculate the possibilities which only one person is sitting next to Clair?(Definitiely this should not include all 3 sitting together case)

Can you please guide me?

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  • $\begingroup$ The number of seating arrangements with all three people together is not $9!3!$ since Clair must be in the middle. $\endgroup$ – N. F. Taussig Aug 25 '17 at 23:34
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Method 1: We use the Inclusion-Exclusion Principle.

Unless otherwise specified, in circular permutations, it is only the relative order that matters.

First, we count the number of distinguishable seating arrangements. We seat Clair. Relative to Clair, the other eleven people can be seated at the table in $11!$ ways as we proceed clockwise around the table.

From these, we must exclude those seating arrangements in which Ann or Ben sit next to Clair.

Arrangements in which Ann sits next to Clair: We treat Ann and Clair as a unit. Since Ann can sit to the left or right of Clair, they can be seated in $2$ ways. Once they are seated, the other ten people can be seated in $10!$ ways as we proceed clockwise around the circle. Hence, there are $2 \cdot 10!$ such seating arrangements.

Arrangements in which Ben sits next to Clair: By the argument given above for Ann and Clair, there are $2 \cdot 10!$ such arrangements.

Arrangements in which both Ann and Ben sit next to Clair: Ann can sit to the left or right of Clair, with Ben occupying the other side, so the trio can be arranged in $2$ ways. The other nine people can be arranged in $9!$ ways as we proceed clockwise around the circle.

By the Inclusion-Exclusion Principle, the number of permissible seating arrangements is $$11! - 2 \cdot 2 \cdot 10! + 2 \cdot 9! = (11 \cdot 10 - 4 \cdot 10 + 2)9! = 72 \cdot 9!$$

Method 2: We correct the solution proposed by pwerth.

We seat Clair. The seat occupied by Clair and the two seats adjacent to Clair cannot be occupied by either Ann or Ben. Hence, there are nine ways to seat Ann and eight ways to seat Ben. The remaining nine people can be seated in $9!$ ways as we proceed clockwise around the circle from Clair. Hence, there are $$9 \cdot 8 \cdot 9!$$ permissible seating arrangements, which agrees with the answer obtained by using the Inclusion-Exclusion Principle.

But how do I calculate the possibilities which only one person is sitting next to Clair?

Method 3: We subtract those arrangements in which exactly one of Ann or Ben sits next to Clair and those arrangements in which both Ann and Ben sit next to Clair from the total.

We showed above that there are $11!$ possible seating arrangements, of which $2 \cdot 9!$ have both Ann and Ben seated next to Clair.

Seating arrangements in which exactly one of Ann or Ben sits next to Clair: Choose whether Ann or Ben sits next to Clair. The chosen person either sits to the left or right of Clair, so Clair and that person can be arranged in two ways. The seat on the other side of Clair cannot be occupied by the person who was not chosen, so that person must sit in one of the nine seats that is not adjacent to Clair. The other nine people can be seated in $9!$ ways as we proceed clockwise around the circle relative to Clair. Hence, there are $2 \cdot 2 \cdot 9 \cdot 9!$ such seating arrangements.

Therefore, the number of permissible seating arrangements is
$$11! - 2 \cdot 2 \cdot 9 \cdot 9! - 2 \cdot 9! = (11 \cdot 10 - 2 \cdot 2 \cdot 9 - 2)9! = 72 \cdot 9!$$

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The problem question "in how many ways can they be seated?" is open to at least two interpretations. In one, the chair positions are numbered, and rotations of any seating arrangement (12 are possible) are to be counted.

This leads to 12*9*8*9! seating arrangements.

Another interpretation, the one the problem author implied by mentioning "ROUND table", is that rotations don't count as unique arrangements. There are 1/12 as many, or

9*8*9!

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  • $\begingroup$ Your answer is incorrect. How did you obtain $10 \cdot 9 \cdot 9!$? $\endgroup$ – N. F. Taussig Aug 25 '17 at 23:14
  • $\begingroup$ I agree, N.F., with your analysis. I made my corrections. $\endgroup$ – pbierre Aug 27 '17 at 3:38

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