How many ways can six people sit around a table, if person 1 isn't next to person 4 and person 5 isn't next to person 6? It's only about the order in which people sit around the table, not which seat they are sitting in. So if all the people will move one seat to the left, it's still the same.
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$\begingroup$ How many ways can you put those people in a line? Now notice how that line becomes a circle as they walk to the table to sit, and what that does to your number of cases. $\endgroup$– The CountJun 16, 2018 at 15:45
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$\begingroup$ I get 24 after screwing around with a pencil for 10 minutes while I wait to go to my wedding (yes I am serious). That is not the final word, but it's what I got. $\endgroup$– The CountJun 16, 2018 at 15:55
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1 Answer
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Assuming that clockwise and anticlockwise sense is different:
We first find out all possible cases without considering the 2 restrictions (1 isn't next to 4 and 5 isn't next to 6) and subtract all the extra cases.
total number of ways of arranging 6 people around a table = 5!
number of ways where 1 and 4 are together = 4!*2
number of ways where 5 and 6 are together = 4!*2
But notice that we have considered the cases where both 1 , 4 and 5 , 6 are together simultaneously 2 times. So we must add that case to get the final answer
number of ways where both 1,4 and 5,6 are together = 3!*2*2
So the final answer is = 5!-4!*2-4!*2+3!*2*2 =48
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1$\begingroup$ nice. i did not assume they were different in my scratch work, which agrees with yours, since i got 24. $\endgroup$ Jul 9, 2018 at 0:47