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we have the following question for homework:

N of the knights of the round table are having dinner. In how many ways can they sit around the table? What if Jack won't sit next to John and the queen - how many possible ways exist now? 

The first question for quite easy - (n-1)!. I'm struggling with the second one: If Jack refuses to sit next to one of them I can count them as one, calculate and then subtract it from the total number of permutations. Its a little harder when he refuses to sit next to two people, because they CAN sit next to each other.

Also, how can I think of it in term of equivalence classes? I am trying to adopt this way of thinking so a hint in that direction would be nice. Thanks in advance!

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    @Angela Richardson: With small typo fix, the solution under first condition (neither next to Jack) is efficient, correct. It solves the problem under the standard interpretation that being the Queen's right-hand man is not the same as being her left-hand man. For second interpretation (want to avoid sitting between) answer is fine if reversal of order means "the same", otherwise need $(N-1)!-2(N-3)!$.2011-11-29

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