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How many different arrangements are there of eight people seated at a round table, where two arrangements are considered the same if one can be obtained from the other by a rotation?

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Imagine that the $8$ people take their seats one at a time, from shortest to tallest. Since orders equivalent by rotation are considered the same, it doesn’t matter where the $A$, first person sits; the arrangement is determined by how the others are seated in relation to $A$. They can then be seated in any order clockwise around the table from $A$’s left. How many permutations of those $7$ people are there?

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Number the seats from one to eight, and then we can try to count the number of ways to put the people around the table, starting by putting someone in seat one, then someone else in seat two, etc. There are eight different people that we could put in the first seat, and then only seven to put in the second, six for the third and so on. This gives us the total number of possible combinations of seatings if we didn't count seatings as the same if they differ only by a rotation. Call this number $n$. But here we have counted different seatings that can be obtained from each other by a rotation several times. In fact, since each seating can be rotated in eight different ways, we have over-counted by a factor of eight, so the answer we want should be $\frac{n}{8}$.

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Since it doesn't matter where the first person sits, the answer would be (8 - 1)! = 7! = 5040

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