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At a dinner party there are 10 people. They all sit at a round table.
In how many ways can they sit if neither Amy nor Lucy want to sit next to Josh?
I have tried this a few times and get a different answer each time. Can you help please?
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In such problems, the indistinguishable arrangements are those that are obtained
one from the other by a circular rotation.
For 10 people around a round table, there are 9! distinguishable arrangements
(= a universal set U of arrangements).
Of them, there are 8! arrangements, where A sits next to Josh on the left of him
and 8! arrangements, where A sits next to Josh on the right of him.
So, in the universal set U, there are 2*8! arrangements {A}, where A sits next to Josh.
Similarly, in the universal set U, there are 2*8! arrangements {L}, where L sits next to Josh.
These arrangements, {A} and {L}, have non-empty intersection.
This intersection has 7! arrangements {AJL} and 7! other arrangements {LJA}.
From it, we conclude that the number of favorable arrangements is
9! - 2*8! - 2*8! + 2*7! = 1*2*3*4*5*6*7*8*9 - 4*(1*2*3*4*5*6*7*8) + 2*(1*2*3*4*5*6*7) = 211680. ANSWER
Solved.
