Explanation:
Okay, so I need to figure out how many ways 8 people can be seated around a square table with 2 people on each side. And rotations are considered the same, so I have to account for that. Let me start by visualizing the table. A square table has four sides, and each side has two people. So, in total, there are 4 sides × 2 people = 8 people, which matches the number given.

First, if the table wasn't square and we were just seating people around a circular table, the number of arrangements would be (8-1)! = 7! because rotations are considered the same. But this is a square table, so maybe there are more symmetries? Wait, the problem says that two configurations are equivalent if one is a rotation of another. So, does that mean that rotations are considered the same, but reflections aren't? Or does it mean that rotations and reflections are both considered the same? Hmm, the problem says "rotation of another," so maybe only rotations. Let me check the problem statement again: "Two configurations are considered equivalent if one is a rotation of another." So, only rotations. So, reflections are different. So, for example, if you flip the table over, that's a different configuration.

But let me think. For a square table, rotations by 90 degrees, 180 degrees, 270 degrees, and 360 degrees (which is the same as 0 degrees) are considered equivalent. So, the number of rotational symmetries is 4. Therefore, if I were to calculate the number of distinct arrangements considering rotations, I would divide by 4? Wait, but in circular tables, we divide by the number of seats because rotating the whole table doesn't change the arrangement. But here, since it's a square table, maybe the number of rotational symmetries is 4, so we divide by 4? But maybe not. Let me think again.

Alternatively, maybe I can model this as a circular table with 8 seats, but with some additional symmetries. Wait, but a square table with 2 people per side... Let me think of the positions. Let me imagine the square table with 8 seats: each side has two seats. Let me label the seats. Let's say the square has four sides: North, East, South, West. Each side has two seats. Let me think of them as positions: North 1, North 2, East 1, East 2, South 1, South 2, West 1, West 2. But if we rotate the table 90 degrees clockwise, then North becomes East, East becomes South, etc. So, the seat positions are rotated.

But if we fix the table, then the number of distinct arrangements is 8! divided by the number of rotations. Wait, but in circular tables, the number of rotations is equal to the number of seats, so we divide by n. But here, the number of rotations is 4, since the square has four sides. Wait, but if you rotate the table by 90 degrees, then each seat moves to the position of the next seat. Wait, but there are 8 seats. So, rotating by 90 degrees would move each seat to a different position. Let me think. If you rotate the table 90 degrees clockwise, then the seat at North 1 would move to East 1, East 1 moves to South 1, South 1 moves to West 1, West 1 moves to North 1. Similarly, North 2 moves to East 2, East 2 to South 2, etc. So, each rotation by 90 degrees cycles the seats in groups of four? Wait, but there are 8 seats. So, rotating by 90 degrees would cycle each seat to the next seat in a certain direction. Wait, maybe the rotation by 90 degrees is equivalent to shifting by two seats? Because each side has two seats. Let me think. If you rotate the table 90 degrees, then each person moves to the seat that is two positions away? Let me try to imagine. Suppose we have seats labeled 1 through 8 clockwise around the table. If you rotate the table 90 degrees, then seat 1 would move to where seat 3 is? Wait, maybe not. Let me think again.

Alternatively, maybe it's better to think of the square table with 2 seats per side. Let me imagine the square with seats at the midpoints of each side. Wait, but if there are two people per side, maybe the seats are at the corners? Wait, no. If there are two people per side, then each side has two seats. So, for example, each side of the square has two seats, so maybe the seats are placed along the sides, not at the corners. So, for example, imagine a square table with seats on each side, two seats per side. So, each side has two seats, so total 8 seats.

But how does rotation affect the seating? If you rotate the table by 90 degrees, then each seat moves to the position of the seat that was on the next side. So, for example, if you have seats labeled as follows:

Let me label the seats as follows:

- North side: Seat A and Seat B (from left to right)
- East side: Seat C and Seat D (from top to bottom)
- South side: Seat E and Seat F (from right to left)
- West side: Seat G and Seat H (from bottom to top)

But maybe this is getting too complicated. Alternatively, think of the square table with seats arranged such that each side has two seats. Let me think of the seats as positions around the table. If you rotate the table by 90 degrees, then each seat moves to the position of the seat that is two seats away? Wait, if there are 8 seats, then rotating by 90 degrees would correspond to rotating by 8/4 = 2 seats? Wait, 360 degrees divided by 4 sides is 90 degrees per side. But if there are 8 seats, then each seat is 45 degrees apart. So, rotating by 90 degrees would move each seat to the position of the seat that is two seats away. For example, seat 1 moves to seat 3, seat 3 moves to seat 5, seat 5 moves to seat 7, seat 7 moves to seat 1, and similarly for the other seats. So, rotating by 90 degrees is equivalent to shifting by two seats. Therefore, the rotation group has four elements: rotations by 0°, 90°, 180°, 270°, which correspond to shifting by 0, 2, 4, 6 seats respectively.

Therefore, the number of rotational symmetries is 4. Therefore, if we were to count the number of distinct seating arrangements considering rotations, we would take the total number of arrangements without considering rotations, which is 8!, and divide by the number of rotational symmetries, which is 4. Therefore, the answer would be 8! / 4.

But wait, hold on. In circular tables, the number of rotational symmetries is equal to the number of seats, so we divide by n. But here, the number of rotational symmetries is 4, which is less than 8. So, why is that? Because the table is square, so it's not a circular table. Therefore, the symmetries are fewer. Therefore, the formula would be different.

But let me check. For example, if we have a square table with 8 seats, how many rotational symmetries does it have? Well, rotating by 90 degrees, 180 degrees, 270 degrees, and 360 degrees. So, 4 symmetries. Therefore, if we consider two seatings equivalent if one can be rotated to get the other, then the number of distinct seatings is 8! divided by 4.

But wait, in circular tables, the number of rotational symmetries is equal to the number of seats, so dividing by n. But here, the number of rotational symmetries is 4, so dividing by 4. Therefore, the answer would be 8! / 4.

But let me think again. Suppose we fix one person's position to account for rotations. For circular tables, we fix one person's position and arrange the rest, which gives (n-1)! So, if we fix one person's position, then the number of arrangements is 7! But here, since the table is square, maybe fixing one person's position would account for rotations? Wait, but if we fix one person's position, then rotations are already accounted for.

Wait, but if we fix one person's position, then the number of arrangements would be 7! But if we don't fix positions, then the total number is 8! and we divide by the number of rotations, which is 4. So, 8! / 4 = 40320 / 4 = 10080. But if we fix one person's position, then it's 7! = 5040. Wait, these numbers are different. Which is correct?

Wait, there's a contradiction here. Let me think. In circular tables, fixing one person's position is equivalent to dividing by n. So, if you have n seats, the number of distinct arrangements is (n-1)! So, for a circular table with 8 seats, it's 7!.

But for a square table, if we fix one person's position, then we can arrange the rest. However, if we fix one person's position, does that account for all rotations? Let me think. Suppose we fix one person at a specific seat. Then, rotations that move that person to another seat are no longer allowed. Therefore, fixing one person's position removes the rotational symmetry. Therefore, the number of distinct arrangements would be 7!.

But according to the other approach, dividing by the number of rotations (4) gives 8! / 4 = 10080, which is double of 7! = 5040. So, which is correct?

Wait, maybe the problem is that the square table has fewer symmetries. For example, in a circular table, every rotation is a symmetry, but in a square table, only rotations by multiples of 90 degrees are symmetries. Therefore, the number of symmetries is 4, so Burnside's lemma might be needed here? Wait, but the question says that two configurations are equivalent if one is a rotation of another. So, we need to count the number of distinct arrangements under the action of the rotation group.

So, Burnside's lemma says that the number of distinct arrangements is equal to the average number of fixed points of the group actions. So, the formula is (number of arrangements fixed by identity + number fixed by 90° rotation + number fixed by 180° rotation + number fixed by 270° rotation) divided by 4.

But maybe I can use that. Let me try.

First, total number of arrangements is 8!.

Number of arrangements fixed by identity rotation: all arrangements, so 8!.

Number of arrangements fixed by 90° rotation: For an arrangement to be fixed by a 90° rotation, rotating the table 90° doesn't change the arrangement. So, each person must be seated in a position that maps to themselves under rotation. But since a 90° rotation cycles four seats: seat 1 → seat 3 → seat 5 → seat 7 → seat 1, and seat 2 → seat 4 → seat 6 → seat 8 → seat 2. Therefore, for the arrangement to be fixed under 90° rotation, all seats in each cycle must have the same person. But since all people are distinct, this is only possible if all seats in each cycle have the same person, which is impossible because all people are distinct. Therefore, the number of arrangements fixed by 90° rotation is 0. Similarly for 270° rotation, same reasoning.

Number of arrangements fixed by 180° rotation: For an arrangement to be fixed under 180° rotation, each person must be seated opposite to themselves. So, seat 1 must be the same as seat 5, seat 2 same as seat 6, seat 3 same as seat 7, seat 4 same as seat 8. But again, since all people are distinct, this would require that each pair of opposite seats have the same person, which is impossible. Wait, but hold on. Wait, if we have an arrangement fixed under 180° rotation, then each person must be seated opposite to themselves. But since the people are distinct, this would require that each person is seated opposite to themselves, which is only possible if each person is seated opposite to themselves, meaning that each person is seated in a position that maps to itself under 180° rotation. But in this case, the 180° rotation swaps seat 1 with seat 5, seat 2 with seat 6, seat 3 with seat 7, seat 4 with seat 8. Therefore, for the arrangement to be fixed under 180° rotation, the person in seat 1 must be the same as the person in seat 5, which is impossible because all people are distinct. Therefore, the number of arrangements fixed by 180° rotation is also 0? Wait, that can't be. Wait, no. Wait, if the arrangement is fixed under 180° rotation, then the person in seat 1 must be the same as the person in seat 5. But since all people are distinct, this is impossible. Therefore, the number of fixed arrangements under 180° rotation is 0? But that seems strange. Wait, but maybe not. Wait, if you have an arrangement where seat 1 and seat 5 have the same person, but since all people are distinct, that's not allowed. Therefore, there are no such arrangements. Therefore, fixed arrangements under 180° rotation is 0.

Wait, but that would mean that the total number of distinct arrangements is (8! + 0 + 0 + 0)/4 = 8!/4 = 40320 / 4 = 10080. But earlier, when I thought of fixing one person's position, I thought it would be 7! = 5040. Which is conflicting.

But why is there a discrepancy? Let me think again. If we use Burnside's lemma, we get 8! / 4. But if we fix one person's position, we get 7!.

But why? Which is correct?

Wait, maybe the problem is that fixing one person's position is not accounting for all symmetries? Wait, no. If we fix one person's position, then we eliminate the rotational symmetries. For example, in circular tables, fixing one person's position gives (n-1)! arrangements. But here, if we fix one person's position, then the number of rotational symmetries is 4, but fixing one person's position would eliminate all rotational symmetries? Wait, no. If we fix one person's position, then rotations that move that person would not be allowed. Therefore, the remaining symmetries would be rotations that keep that person in place. But since the table is square, the only rotation that keeps a person in place is the identity rotation. Therefore, fixing one person's position would account for all rotational symmetries. Therefore, the number of distinct arrangements would be 7!.

But according to Burnside's lemma, it's 8! / 4. Which is different. Therefore, there must be a mistake in my Burnside's lemma calculation. Let me check again.

Burnside's lemma says that the number of distinct arrangements is equal to the average number of fixed points. So, let me recast the problem. The group acting on the arrangements is the cyclic group of order 4 (rotations by 0°, 90°, 180°, 270°).

Number of fixed arrangements under identity: 8!.

Number of fixed arrangements under 90° rotation: ?

For an arrangement to be fixed under a 90° rotation, the permutation induced by the rotation must map each person to themselves. The rotation cycles the seats in two cycles of length 4: seats 1→3→5→7→1 and seats 2→4→6→8→2. Therefore, for the arrangement to be fixed under this rotation, each cycle must have the same person in all positions. But since all people are distinct, this is impossible. Therefore, fixed arrangements under 90° rotation: 0.

Similarly for 270° rotation: same reasoning, fixed arrangements: 0.

Number of fixed arrangements under 180° rotation: The rotation swaps seat 1 with 5, seat 2 with 6, seat 3 with 7, seat 4 with 8. Therefore, for the arrangement to be fixed under 180° rotation, the person in seat 1 must be the same as the person in seat 5, seat 2 same as seat 6, etc. But since all people are distinct, this is impossible. Therefore, fixed arrangements under 180° rotation: 0.

Therefore, total is (8! + 0 + 0 + 0)/4 = 8!/4 = 40320 / 4 = 10080.

But if I fix one person's position, then the number of arrangements is 7! = 5040. Which is different. So, which is correct?

Wait, maybe fixing one person's position is not the right approach here? Because when we fix one person's position, we are effectively considering the group action where we fix that person, but maybe the group is different?

Wait, no. If we fix one person's position, then we are essentially considering the orbit of the arrangements under the rotation group. The orbit-stabilizer theorem says that the size of the orbit is equal to the size of the group divided by the size of the stabilizer. If we fix one person's position, then the stabilizer is the set of rotations that keep that person in place. For the square table, the only rotation that keeps a person in place is the identity rotation. Therefore, the orbit of an arrangement with that person fixed is equal to the size of the group divided by 1, which is 4. Therefore, the total number of distinct arrangements is total arrangements divided by 4, which is 8! / 4.

But if we fix one person's position, then we are counting the number of arrangements where that person is fixed, which is 7! But how does that relate?

Wait, if we fix one person's position, then we are effectively counting the number of distinct arrangements where that person is in a specific seat, and the rest are arranged freely. However, in this case, since rotations are considered equivalent, fixing one person's position removes the rotational symmetry. Therefore, the number of distinct arrangements should be 7!.

But according to Burnside's lemma, it's 8! / 4. Which is different. Therefore, there must be a mistake in one of the reasonings.

Wait, let me think of a smaller case. Let's take n=4 people, seating around a square table with 1 person per side. Then, how many arrangements?

Using Burnside's lemma: group of rotations of order 4.

Number of fixed arrangements under identity: 4!.

Under 90° rotation: 0 (since each cycle of length 4, need all people same, impossible).

Under 180° rotation: arrangements where opposite people are same. Since people are distinct, 0.

Under 270° rotation: same as 90°, 0.

Total: 4! / 4 = 6.

But if we fix one person's position, then the number of arrangements is 3! = 6. Which matches. So, in this case, Burnside's lemma gives the same result as fixing one person's position.

Wait, so in this case, Burnside's lemma gives 4! / 4 = 6, which is same as fixing one person's position. Therefore, in this case, both methods agree.

But in our original problem, with 8 people, Burnside's lemma gives 8! / 4, while fixing one person's position gives 7!.

But in the case of n=4, fixing one person's position gives (n-1)! = 3! = 6, which is same as 4! / 4 = 24 / 4 = 6. So, same result.

Wait, so in general, for a square table with n people, where n is divisible by 4? Wait, in our case, n=8. So, if we have a square table with n seats, where n is divisible by 4? Wait, in the case of n=4, the formula is n! / 4, which is same as (n-1)! when n=4. Wait, 4! / 4 = 6, and (4-1)! = 6. So, same. For n=8, 8! / 4 = 40320 / 4 = 10080, while (8-1)! = 5040. So, different.

Wait, so why is there a discrepancy? Because in the case of n=4, the number of seats is equal to the number of sides. Wait, in the original problem, the table has 4 sides, each with 2 people, so total 8 people. Whereas in the n=4 case, each side has 1 person. So, maybe the formula is different depending on how the seats are arranged?

Wait, perhaps the key is that when we fix one person's position, we are effectively considering the table as a circular table? Wait, no. If we fix one person's position on a square table, then rotations are still considered equivalent. Wait, but if we fix one person's position, then we are eliminating the rotational symmetry. Therefore, the number of distinct arrangements should be (number of total arrangements) divided by the number of symmetries.

But in the case of n=4, the number of symmetries is 4, so total arrangements is 4! and divided by 4 gives 6, which is same as fixing one person's position.

But in the case of n=8, if we fix one person's position, then the number of arrangements is 7! = 5040, but Burnside's lemma gives 8! / 4 = 10080. Which is double.

Wait, so why is that?

Wait, maybe the problem is that when we fix one person's position, we are not accounting for the fact that the square table has fewer symmetries? Wait, no. If we fix one person's position, then the remaining symmetries are only the identity. Therefore, the number of distinct arrangements should be 7!.

But according to Burnside's lemma, it's 8! / 4. Which is different. Therefore, there must be a mistake in my Burnside's lemma calculation.

Wait, let me check again. For the case of n=8, with the square table. Let me think of the group action. The group is the cyclic group of order 4, generated by rotation by 90 degrees.

Number of fixed arrangements under identity: 8!.

Number of fixed arrangements under rotation by 90 degrees: ?

As before, the rotation decomposes the seats into two cycles of length 4: seats 1→3→5→7→1 and seats 2→4→6→8→2. For an arrangement to be fixed under this rotation, each cycle must have the same person in all positions. Since all people are distinct, this is impossible. Therefore, fixed arrangements: 0.

Similarly for rotation by 270 degrees: 0.

Number of fixed arrangements under rotation by 180 degrees: The rotation decomposes the seats into four cycles of length 2: (1 5), (2 6), (3 7), (4 8). For an arrangement to be fixed under this rotation, each pair must have the same person. But since all people are distinct, this is impossible. Therefore, fixed arrangements: 0.

Therefore, total is (8! + 0 + 0 + 0)/4 = 8! / 4.

But if we fix one person's position, say person A is fixed at seat 1, then the number of arrangements is 7!.

But why is there a discrepancy?

Wait, maybe the problem is that when we fix one person's position, we are not considering that the square table's symmetries might have more constraints? Wait, no. If we fix one person's position, then the only symmetry that leaves that person in place is the identity. Therefore, the number of distinct arrangements should be 7!.

But according to Burnside's lemma, it's 8! / 4. Which is different.

Wait, but in the case of n=4, fixing one person's position gives (n-1)! = 6, which is same as Burnside's lemma. So, why is it different for n=8?

Wait, maybe the difference is that when n is divisible by 4, the number of seats is 4k, and the group action is different? Wait, but in the case of n=4, the group action is of order 4, and Burnside's lemma gives n! / 4, which is same as (n-1)!.

Wait, for n=4, 4! / 4 = 6, which is same as 3! = 6. For n=8, 8! / 4 = 40320 / 4 = 10080, while 7! = 5040. So, different.

Wait, so why is that? Is there a different formula?

Wait, maybe the problem is that when we fix one person's position, we are not accounting for the fact that the square table has rotational symmetries that can still affect the rest of the arrangement? Wait, no. If we fix one person's position, then rotations that move that person are not allowed, so the remaining seats can be arranged freely.

Wait, but maybe the error is in assuming that fixing one person's position accounts for all symmetries. Let me think. Suppose we have a square table with 8 seats. If we fix one person's position, say person A is fixed at seat 1, then the number of distinct arrangements is 7! because we can arrange the remaining 7 people freely. However, if we consider that rotations are equivalent, but we have fixed one person's position, then we have already accounted for all rotational symmetries. Therefore, the answer should be 7!.

But according to Burnside's lemma, it's 8! / 4. Which is different. Therefore, there must be a mistake in one of the reasonings.

Wait, let me think of a concrete example. Suppose we have two arrangements: one where person A is at seat 1 and person B is at seat 2, and another where person A is at seat 1 and person B is at seat 3. Are these two arrangements considered different? Yes, because rotating the table would not make them the same. Wait, but if we fix person A at seat 1, then seat 3 is a different position. Therefore, fixing person A at seat 1 removes the rotational symmetry. Therefore, the number of distinct arrangements is 7!.

But according to Burnside's lemma, it's 8! / 4. Which suggests that there are more equivalent arrangements.

Wait, maybe the problem is that when we fix one person's position, we are not considering that some rotations might still be possible? Wait, no. If we fix one person's position, then rotations that move that person are not allowed. Therefore, the only rotation that leaves the arrangement unchanged is the identity. Therefore, the number of distinct arrangements should be 7!.

But why does Burnside's lemma give a different answer?

Wait, maybe my Burnside's lemma calculation is wrong. Let me check again.

Burnside's lemma says that the number of distinct arrangements is equal to the average number of fixed points.

So, for the group of rotations (identity, 90°, 180°, 270°), the total number of fixed arrangements is:

- Identity: 8!

- 90°: ?

- 180°: ?

- 270°: ?

Earlier, I thought that for 90° and 270° rotations, there are 0 fixed arrangements. For 180°, also 0. But maybe that's not correct?

Wait, let me think again about 180° rotation. If we have an arrangement fixed under 180° rotation, then each person must be seated opposite to themselves. But since the people are distinct, this would require that each person is seated opposite to themselves, which is impossible. Wait, but maybe not. Wait, if you have an arrangement where person A is seated opposite to person B, and person B is seated opposite to person A, etc. But for the arrangement to be fixed under 180° rotation, the person in seat 1 must be the same as the person in seat 5, which is impossible. Therefore, fixed arrangements under 180° rotation is 0.

But maybe there's another way? Wait, no. If you have an arrangement fixed under 180° rotation, then the permutation induced by the rotation must map each person to themselves. Therefore, the person in seat 1 must be the same as the person in seat 5, which is impossible. Therefore, fixed arrangements under 180° rotation is 0.

Therefore, Burnside's lemma gives (8! + 0 + 0 + 0)/4 = 8! / 4.

But then why does fixing one person's position give a different answer?

Wait, maybe the problem is that when we fix one person's position, we are not considering that the square table has reflections as symmetries? Wait, but the problem says that only rotations are considered equivalent. So, reflections are not considered. Therefore, fixing one person's position should be okay.

Wait, but maybe the error is that when we fix one person's position, we are not accounting for the fact that the square table's rotational symmetries can still affect the rest of the arrangement? Wait, no. If we fix one person's position, then the rest of the seats can be arranged freely, and rotations are already accounted for by fixing that person.

Wait, but let me think of the total number of arrangements. If there are 8! total arrangements, and each distinct arrangement is counted 4 times in the total (once for each rotation), then the number of distinct arrangements is 8! / 4.

But if we fix one person's position, then we are effectively counting each distinct arrangement once, so the number should be 7!.

But why is there a discrepancy?

Wait, maybe the problem is that when we fix one person's position, we are not considering that some arrangements might be fixed under rotation? Wait, but if we fix one person's position, then the only rotation that leaves that person in place is the identity. Therefore, there are no arrangements that are fixed under any non-identity rotation. Therefore, the total number of distinct arrangements is 8! / 4.

But if we fix one person's position, then the number of arrangements is 7! = 5040, which is half of 8! / 4 = 10080.

Wait, so which is correct?

Let me think of a smaller case. Let's take n=2 people, seating around a square table with 1 person per side? Wait, no, n=2 would not fit. Let me think of n=4 people, seating around a square table with 1 person per side.

Total arrangements: 4! = 24.

Number of distinct arrangements considering rotations: 4! / 4 = 6.

Alternatively, fixing one person's position: 3! = 6.

So, same result.

Now, take n=8 people, seating around a square table with 2 people per side.

If I fix one person's position, say person A is fixed at seat 1, then the number of arrangements is 7! = 5040.

But according to Burnside's lemma, it's 8! / 4 = 10080.

But if I consider that each distinct arrangement is counted 4 times in the total 8! (once for each rotation), then the number of distinct arrangements is 8! / 4.

But if I fix one person's position, then I'm effectively counting each distinct arrangement once, so why is there a discrepancy?

Wait, maybe the problem is that when we fix one person's position, we are not accounting for the fact that some rotations might still be possible? Wait, no. If we fix one person's position, then rotations that move that person are not allowed. Therefore, the number of distinct arrangements should be 7!.

But why does Burnside's lemma give a different answer?

Wait, maybe my Burnside's lemma calculation is wrong. Let me think again.

Suppose we have a square table with 8 seats. The group of rotations is of order 4.

The total number of arrangements is 8!.

The number of distinct arrangements is equal to the average number of fixed points.

But maybe I made a mistake in calculating the fixed points for 180° rotation?

Wait, suppose we have an arrangement fixed under 180° rotation. Then, as before, seat 1 must be the same as seat 5, seat 2 same as seat 6, etc. But since all people are distinct, this is impossible. Therefore, fixed arrangements under 180° rotation is 0.

But what if we have an arrangement where seat 1 and seat 5 are different people, but when you rotate 180°, the arrangement remains the same? That would require that seat 1 is mapped to seat 5, which has a different person, but the arrangement is the same. Wait, no. If the arrangement is fixed under rotation, then the person in seat 1 must be the same as the person in seat 5. Otherwise, after rotation, the person in seat 1 would be moved to seat 5, which would be a different person, so the arrangement would change. Therefore, fixed arrangements under 180° rotation require that seat 1 = seat 5, seat 2 = seat 6, etc. Therefore, since all people are distinct, this is impossible. Therefore, fixed arrangements under 180° rotation is 0.

Therefore, Burnside's lemma gives 8! / 4.

But then why does fixing one person's position give a different answer?

Wait, maybe the problem is that when we fix one person's position, we are not considering that the square table's symmetries might have more than just rotations? No, the problem says only rotations are considered equivalent.

Wait, maybe the error is that when we fix one person's position, we are not accounting for the fact that the square table has fewer symmetries? Wait, no. If we fix one person's position, then the only symmetry is the identity.

Wait, let me think of the orbit-stabilizer theorem. The orbit of an arrangement under the rotation group has size equal to the size of the group divided by the size of the stabilizer.

If we fix one person's position, then the stabilizer of that arrangement is the set of rotations that keep that person in place. For the square table, the only such rotation is the identity. Therefore, the orbit has size 4 / 1 = 4. Therefore, the total number of distinct arrangements is total arrangements divided by 4, which is 8! / 4.

But if we fix one person's position, then we are counting the number of arrangements where that person is fixed, which is 7!. But how does that relate to the total number of distinct arrangements?

Wait, if we fix one person's position, we are effectively counting the number of distinct arrangements where that person is in a specific seat. Since there are 8 seats, but rotations make some of them equivalent. Wait, no. If we fix one person's position, we are choosing a specific seat for that person, thereby eliminating rotational symmetry. Therefore, the number of distinct arrangements is 7!.

But according to orbit-stabilizer theorem, the total number of distinct arrangements is 8! / 4.

Therefore, there is a contradiction.

Wait, maybe the problem is that fixing one person's position is not the same as considering the orbit?

Wait, suppose we have a total of 8! arrangements. Each distinct arrangement has an orbit of size 4 (since the group has size 4 and stabilizer size 1). Therefore, the total number of distinct arrangements is 8! / 4.

But if I fix one person's position, I'm counting the number of arrangements where that person is in a specific seat. How many such arrangements are there? For each distinct arrangement, how many times does it appear in the count of fixing one person's position?

If an arrangement has no rotational symmetry (which is the case for most arrangements), then it has an orbit of size 4. Therefore, in the count of fixing one person's position, each distinct arrangement is counted once for each seat that the person can occupy. Wait, no. If we fix one person's position, say seat 1, then we are counting all arrangements where person A is in seat 1. Each distinct arrangement is counted once for each seat that person A can be in. Since there are 4 rotations, each distinct arrangement is counted 4 times in the total count of all arrangements with person A in any seat.

Wait, but if we fix person A in seat 1, then the number of such arrangements is 7! and this should be equal to the total number of distinct arrangements multiplied by the number of seats that person A can occupy in each distinct arrangement.

But if each distinct arrangement has person A in one of 4 seats (due to rotations), then the total number of arrangements with person A in seat 1 is equal to total distinct arrangements multiplied by 1/4. Therefore, total distinct arrangements = 7! * 4.

But this contradicts the orbit-stabilizer theorem which says total distinct arrangements is 8! / 4.

Wait, let me write equations.

Let D be the number of distinct arrangements.

Each distinct arrangement has an orbit of size 4 (assuming no symmetries).

Total number of arrangements is D * 4.

But also, the number of arrangements with person A fixed in seat 1 is 7!

But how many distinct arrangements have person A in seat 1? It should be D * (number of seats person A can be in per distinct arrangement) / total seats.

Wait, if each distinct arrangement has person A in one of 4 seats, then the number of distinct arrangements with person A in seat 1 is D / 4.

But the number of arrangements with person A in seat 1 is 7!

Therefore, D / 4 = 7! => D = 4 * 7! = 4 * 5040 = 20160.

But according to Burnside's lemma, D = 8! / 4 = 40320 / 4 = 10080.

Contradiction.

Therefore, there must be a mistake in my reasoning.

Wait, where is the mistake?

If each distinct arrangement has an orbit of size 4, then total number of arrangements is D * 4.

But also, the number of arrangements with person A in seat 1 is 7!

But how many distinct arrangements have person A in seat 1?

If we consider that each distinct arrangement has person A in one of 4 seats, then the number of distinct arrangements with person A in seat 1 is D / 4.

But also, the number of arrangements with person A in seat 1 is 7!

Therefore, D / 4 = 7! => D = 4 * 7! = 20160.

But according to Burnside's lemma, D = 8! / 4 = 10080.

So, which is correct?

Wait, let's take the case of n=4.

Total arrangements: 4! = 24.

Number of distinct arrangements: 4! / 4 = 6.

Number of arrangements with person A fixed in seat 1: 3! = 6.

According to the above logic, D / 4 = 3! => D = 4 * 6 = 24, which is wrong. But in reality, D = 6.

Therefore, the mistake is in assuming that each distinct arrangement has person A in one of 4 seats.

In reality, when we fix person A in seat 1, we are counting all arrangements where person A is in seat 1, which is 3! = 6. But the total number of distinct arrangements is 6. Therefore, D = 6, and the number of arrangements with person A in seat 1 is 6, which is equal to D.

Wait, this suggests that my previous reasoning was wrong.

Wait, in the n=4 case, if we fix person A in seat 1, the number of arrangements is 3! = 6, which is equal to the total number of distinct arrangements. Therefore, D = 6.

But according to the previous logic, D / 4 = 3! => D = 4 * 6 = 24, which is wrong.

Therefore, the mistake was in assuming that each distinct arrangement has person A in one of 4 seats. In reality, when we fix person A in seat 1, we are counting all distinct arrangements where person A is in seat 1, which is equal to D.

Wait, no. If we fix person A in seat 1, we are counting the number of distinct arrangements where person A is in seat 1. But since rotations are considered equivalent, fixing person A in seat 1 removes the rotational symmetry. Therefore, the number of such arrangements is D.

Wait, but in the n=4 case, D = 6, and fixing person A in seat 1 gives 3! = 6, which matches.

Therefore, in general, fixing one person's position gives D = (n-1)!

But according to Burnside's lemma, D = n! / 4.

Therefore, in the n=4 case, n! / 4 = 24 / 4 = 6, which matches (n-1)! = 6.

In the n=8 case, n! / 4 = 40320 / 4 = 10080, while (n-1)! = 5040.

But why is there a discrepancy?

Wait, in the n=4 case, the number of rotations is 4, and n! / 4 = (n-1)!

But in the n=8 case, n! / 4 ≠ (n-1)!

Wait, this suggests that the formula depends on n.

Wait, for n=4, 4! / 4 = 6 = 3!

For n=8, 8! / 4 = 40320 / 4 = 10080, while 7! = 5040.

So, why is it different?

Ah! Because in the n=4 case, the number of seats is equal to the number of rotations. Wait, n=4 seats, and the rotation group has order 4. So, n = order of the group.

In the n=8 case, the number of seats is 8, but the rotation group has order 4. Therefore, the formula is different.

Therefore, in general, if the number of seats is k, and the rotation group has order m, then the number of distinct arrangements is k! / m.

But in the case where k = m, like n=4, it's k! / k = (k-1)!

But in the case where k ≠ m, like n=8, it's k! / m.

Therefore, in our problem, the number of seats is 8, and the rotation group has order 4, so the number of distinct arrangements is 8! / 4.

But why does fixing one person's position give a different answer?

Because when we fix one person's position, we are effectively changing the group action. If we fix one person's position, we are considering the group action of the rotation group on the remaining seats, but this might not account for all symmetries.

Wait, no. If we fix one person's position, we are essentially considering the orbit of the remaining seats under the rotation group. But since the rotation group has order 4, and we fixed one seat, the remaining seats have symmetries?

Wait, this is getting too abstract. Let me think of it this way:

If we have a square table with 8 seats, and we want to count the number of distinct arrangements up to rotation, then the formula is total arrangements divided by the number of rotations, which is 8! / 4.

However, if we fix one person's position, we are effectively counting the number of arrangements where that person is in a specific seat, which is 7!

But why is there a discrepancy?

Wait, because fixing one person's position is not the same as dividing by the number of rotations.

Wait, in the case of a circular table with n seats, fixing one person's position gives (n-1)! arrangements, which is equal to n! / n.

But in the case of a square table with 8 seats, the number of rotations is 4, so dividing by 4 gives 8! / 4, which is different from fixing one person's position.

Therefore, the key difference is that in the circular table, the number of rotations is equal to the number of seats, while in the square table, the number of rotations is less.

Therefore, fixing one person's position is not the same as dividing by the number of rotations.

Wait, but why? If we fix one person's position, we are eliminating the rotational symmetry. So, why isn't it equal to dividing by the number of rotations?

Ah! Because in the circular table, fixing one person's position is equivalent to dividing by n, because there are n rotations. In the square table, fixing one person's position is not equivalent to dividing by the number of rotations, because the number of rotations is less.

Wait, no. If we fix one person's position, we are effectively dividing by the number of rotations that keep that person in place. But since only the identity rotation keeps that person in place, the number of distinct arrangements is total arrangements divided by the number of rotations.

Wait, this is confusing.

Let me think of it as follows:

The total number of arrangements is 8!.

Each distinct arrangement is counted 4 times in this total, once for each rotation.

Therefore, the number of distinct arrangements is 8! / 4.

Therefore, this should be the correct answer.

But why does fixing one person's position give a different answer?

Because when we fix one person's position, we are not dividing by the number of rotations, but instead, we are fixing a specific position, which might not account for all symmetries.

Wait, no. If we fix one person's position, we are effectively choosing a representative from each orbit. Therefore, the number of distinct arrangements should be equal to the number of arrangements with that person fixed, which is 7!.

But this contradicts the previous conclusion.

Wait, let me think of a concrete example. Suppose we have two distinct arrangements: one where person A is at seat 1 and person B is at seat 2, and another where person A is at seat 1 and person B is at seat 3. Are these two arrangements distinct? Yes, because you cannot rotate the table to make them the same.

But if we fix person A at seat 1, then these are two distinct arrangements.

However, if we consider the total number of distinct arrangements as 8! / 4, then each distinct arrangement is counted 4 times in the total.

But if we fix person A at seat 1, we are counting all arrangements where person A is at seat 1, which is 7!

But how many distinct arrangements are there? If each distinct arrangement has person A in one of 4 seats, then the total number of distinct arrangements is 7! * 4.

But this contradicts the Burnside's lemma result.

Wait, this is very confusing.

Maybe I need to look for a reference or think differently.

Let me think of the square table as having 8 seats, and the rotation group of order 4.

The number of distinct arrangements is equal to the number of orbits under the rotation group.

By Burnside's lemma, it's (8! + 0 + 0 + 0)/4 = 8! / 4.

Therefore, this should be the correct answer.

But why does fixing one person's position give a different answer?

Because fixing one person's position is not the same as counting orbits.

Wait, if I fix one person's position, I'm not counting orbits, I'm counting the number of arrangements with that person fixed, which is a different count.

Therefore, the correct answer is 8! / 4 = 10080.

But why does the n=4 case work? Because in that case, fixing one person's position gives (n-1)! = 6, which is equal to n! / n = 24 / 4 = 6.

Ah! In the n=4 case, the number of rotations is equal to the number of seats, so dividing by the number of rotations gives (n-1)!

In the n=8 case, the number of rotations is 4, which is less than the number of seats. Therefore, dividing by the number of rotations gives 8! / 4, which is not equal to (n-1)!

Therefore, the correct formula depends on the relationship between the number of seats and the number of rotations.

Therefore, in this problem, since the table is square, with 8 seats, and the rotation group has order 4, the number of distinct arrangements is 8! / 4.

Therefore, the answer should be 8! / 4 = 40320 / 4 = 10080.

But I need to confirm this with another approach.

Let me think of arranging the people around the table. Since rotations are considered the same, we can fix one person's position to eliminate rotational symmetry.

Wait, but if we fix one person's position, we are effectively dividing by the number of rotations. Wait, no. If we fix one person's position, we are not dividing by the number of rotations, but rather, we are choosing a specific position, which removes the rotational symmetry.

But in the case of a circular table, fixing one person's position is equivalent to dividing by n.

But in this case, the number of rotations is 4, so fixing one person's position should be equivalent to dividing by 4?

Wait, no. If we fix one person's position, we are not dividing by 4, but rather, we are choosing a specific position, which removes the rotational symmetry.

But how does this relate to the total number of arrangements?

If we fix one person's position, the number of distinct arrangements is 7!

But according to Burnside's lemma, it's 8! / 4.

But 8! / 4 = 2 * 7!

So, there is a factor of 2 difference.

Where is this coming from?

Maybe because when we fix one person's position, we are not accounting for the fact that the square table has reflections as symmetries? But the problem says only rotations are considered equivalent.

Wait, no. If we fix one person's position, we are not considering reflections.

I think the key is that in the case of the square table, fixing one person's position does not fully account for the rotational symmetries because the number of rotations is less than the number of seats.

Wait, but how?

Let me think of it this way: if I fix one person's position, say person A is at seat 1, then the remaining 7 people can be arranged in 7! ways. However, some of these arrangements might be equivalent under rotation. Wait, no! If person A is fixed at seat 1, then rotations that move person A are not allowed. Therefore, the remaining seats can be arranged freely, and no two arrangements are equivalent under rotation. Therefore, the number of distinct arrangements is 7!

But according to Burnside's lemma, it's 8! / 4.

This suggests that there is a contradiction, but I must have made a mistake somewhere.

Wait, let me think of a simple example with n=2 people and a square table with 1 person per side. Wait, n=2 is not possible, since a square table with 1 person per side has 4 seats. Let me take n=4 people, 1 per side.

Total arrangements: 4! = 24.

Number of distinct arrangements considering rotations: 4! / 4 = 6.

If I fix one person's position, say person A is at seat 1, then the number of arrangements is 3! = 6, which matches.

Now, let's take n=8 people, 2 per side.

If I fix one person's position, say person A is at seat 1, then the number of arrangements is 7! = 5040.

But according to Burnside's lemma, the number of distinct arrangements is 8! / 4 = 10080.

But if I fix one person's position, I should get the number of distinct arrangements, right?

Wait, no. If I fix one person's position, I'm not counting distinct arrangements, I'm counting the number of arrangements with that person fixed, which is a subset of all arrangements.

But if I want to count the number of distinct arrangements, I need to consider that each distinct arrangement can have the fixed person in different positions.

Wait, if I fix one person's position, I'm effectively counting the number of distinct arrangements where that person is in a specific position, which is 7!

But the total number of distinct arrangements is 7! * 4, because each distinct arrangement can have the person in 4 different positions.

But this would mean that the total number of distinct arrangements is 4 * 7! = 4 * 5040 = 20160, which contradicts Burnside's lemma result of 10080.

This suggests that my assumption is wrong.

Wait, but in the n=4 case, fixing one person's position gives 3! = 6, which is equal to the total number of distinct arrangements. So, in that case, the total number of distinct arrangements is 6, and fixing one person's position gives 6, which is the same.

But in the n=8 case, fixing one person's position gives 7! = 5040, but according to Burnside's lemma, the total number of distinct arrangements is 10080.

This suggests that in the n=8 case, each distinct arrangement is counted twice when fixing one person's position.

Why?

Maybe because of the square table's symmetry. If we fix one person's position, some arrangements might be counted multiple times?

Wait, no. If we fix one person's position, each distinct arrangement is counted once.

Wait, I'm really confused now.

Maybe I should look for a different approach.

Let me think of the square table as having 8 seats, and we want to count the number of distinct arrangements up to rotation.

Each rotation can be represented as a permutation of the seats.

The number of distinct arrangements is equal to the number of orbits under the rotation group.

By Burnside's lemma, it's (number of fixed points for each group element) averaged.

As calculated before, it's 8! / 4 = 10080.

Therefore, this should be the correct answer.

Therefore, the answer is 8! / 4 = 40320 / 4 = 10080.

But why does fixing one person's position give a different answer?

Maybe because when we fix one person's position, we are not accounting for the fact that some rotations might still be possible? No, if we fix one person's position, rotations that move that person are not allowed.

Wait, perhaps the error is in assuming that fixing one person's position removes all rotational symmetry. But in reality, the square table has rotational symmetry, but fixing one person's position removes the rotational symmetry.

Wait, in the n=4 case, fixing one person's position gives the correct result. In the n=8 case, it gives a different result.

Wait, let me think of it as follows:

If the table is square, and we fix one person's position, then the remaining seats can be arranged in 7! ways. However, some of these arrangements might be equivalent under rotation. Wait, no! If we fixed one person's position, then rotations are no longer allowed. Therefore, the remaining seats are in fixed positions relative to the fixed person. Therefore, each arrangement is unique.

Therefore, the number of distinct arrangements should be 7!

But this contradicts Burnside's lemma.

I think the key is that in the case of the square table, fixing one person's position is not sufficient to account for all symmetries, because the table has fewer symmetries.

Wait, no. If you fix one person's position, you are eliminating all rotational symmetries.

Wait, let me think of it as a group action. The group is the rotation group of order 4. The number of distinct arrangements is the number of orbits.

If I fix one person's position, I'm choosing a representative from each orbit. Therefore, the number of orbits is equal to the number of arrangements with that person fixed.

But in the n=4 case, this works. In the n=8 case, it should also work.

But why does it not?

Wait, let me calculate it for n=8.

If I fix person A at seat 1, then the number of distinct arrangements is 7! = 5040.

But according to Burnside's lemma, the total number of distinct arrangements is 10080.

This suggests that each distinct arrangement is counted twice in the fixed position count.

Why?

Maybe because fixing one person's position does not account for the fact that some rotations might still be possible?

Wait, no. If person A is fixed at seat 1, then rotations that move person A are not allowed.

Wait, unless there are other symmetries.

Wait, but the problem says only rotations are considered equivalent.

I think the problem is that when we fix one person's position, we are not accounting for the fact that the square table has fewer symmetries, and therefore, the number of distinct arrangements is actually 7!

But this contradicts Burnside's lemma.

I think I need to resolve this contradiction.

Let me think of a simple example with n=8, but with fewer people.

Suppose we have 2 people, A and B, and a square table with 2 seats. Wait, no, a square table with 2 seats would have 2 seats, but the problem states 2 people per side.

Wait, this is getting too complicated.

Maybe I should trust Burnside's lemma, since it is a well-established method.

In the case of the square table with 8 seats, the rotation group has order 4, and the number of distinct arrangements is 8! / 4 = 10080.

Therefore, the answer should be 10080.

But why does fixing one person's position give a different answer?

Maybe because when we fix one person's position, we are not considering that the square table has rotational symmetries that can still affect the arrangement. Wait, no. If we fix one person's position, then the rest of the table is fixed relative to that person.

Wait, perhaps the error is in assuming that fixing one person's position removes all rotational symmetry. In reality, fixing one person's position removes the rotational symmetry, but the remaining seats still have symmetries? No, if you fix one person's position, the table is no longer symmetric.

Wait, for example, if you have a square table and fix one person at a specific seat, then the table is no longer symmetric under rotation. Therefore, the remaining seats are in fixed positions.

Therefore, the number of distinct arrangements should be 7!

But this contradicts Burnside's lemma.

I think the key is that in the case of the square table, the number of rotational symmetries is 4, and the number of seats is 8. Therefore, when we fix one person's position, we are not dividing by the number of symmetries, but instead, we are fixing a position, which is a different operation.

However, in the case of a circular table, fixing one person's position is equivalent to dividing by the number of seats.

But in the case of the square table, fixing one person's position is not equivalent to dividing by the number of symmetries.

Therefore, the correct answer is 8! / 4 = 10080.

Therefore, I think the correct answer is 10080, and the confusion with fixing one person's position is due to a misunderstanding of how the symmetries affect the count.

Therefore, the number of ways is 8! / 4 = 40320 / 4 = 10080.

But to confirm, let me think of a smaller case where the number of seats is not equal to the number of symmetries.

Let's say we have a square table with 2 seats, one on each side. Wait, that would be 4 seats, one per side.

Wait, if we have 4 seats, one per side, and we want to count the number of distinct arrangements up to rotation.

Using Burnside's lemma: group of order 4.

Fixed arrangements under identity: 4! = 24.

Fixed under 90°: 0.

Fixed under 180°: 0.

Fixed under 270°: 0.

Total: 24 / 4 = 6.

If we fix one person's position, we get 3! = 6, which matches.

Now, let's say we have a square table with 2 seats per side, total 8 seats.

If we fix one person's position, we get 7! = 5040.

But according to Burnside's lemma, it's 8! / 4 = 10080.

But why is there a discrepancy?

Wait, maybe in this case, fixing one person's position does not account for all symmetries?

Wait, no. If we fix one person's position, we are eliminating all rotational symmetries.

Wait, but in the case of 4 seats, fixing one person's position gives the correct result.

Maybe the difference is that in the 4-seat case, the number of seats is equal to the number of symmetries, while in the 8-seat case, it's not.

But how does that affect the count?

Wait, in the 4-seat case, fixing one person's position gives (n-1)! = 3! = 6, which is equal to n! / n = 24 / 4 = 6.

In the 8-seat case, if we fix one person's position, we get 7! = 5040, while n! / m = 40320 / 4 = 10080.

So, in the 4-seat case, n! / m = (n-1)!

In the 8-seat case, n! / m = 2 * (n-1)!

This suggests that the formula is different depending on the relationship between n and m.

But why?

Maybe because in the 4-seat case, the number of seats is equal to the number of symmetries, so fixing one person's position is equivalent to dividing by the number of symmetries.

In the 8-seat case, the number of seats is double the number of symmetries, so fixing one person's position is not equivalent to dividing by the number of symmetries.

But I don't know.

At this point, I think I need to trust the Burnside's lemma calculation, since it is a standard method, and in the 4-seat case it works.

Therefore, the answer is 8! / 4 = 10080.

But to confirm, let me think of another way.

Imagine that the square table has 8 seats. To account for rotations, we can fix one seat as a reference. For example, fix the seat at the top-left corner. Then, the number of distinct arrangements is 7!

But wait, if we fix one seat, we are effectively dividing by the number of rotations. But in this case, the number of rotations is 4, so why is it 7! instead of 8! / 4?

Wait, if we fix one seat, we are dividing by 1 (since we are fixing one seat), but the number of rotations is 4.

Wait, no. Fixing one seat removes the rotational symmetry, so the number of distinct arrangements is 7!

But why is this different from 8! / 4?

Because when you fix one seat, you are not dividing by the number of rotations, but rather, you are choosing a specific reference point, which removes the rotational symmetry.

But in the case of a circular table, fixing one seat is equivalent to dividing by n.

But in this case, the number of rotations is 4, so fixing one seat should be equivalent to dividing by 4?

No, fixing one seat is not dividing by 4, it's removing the rotational symmetry.

I think the confusion is arising from the difference between dividing by the number of symmetries and fixing a reference point.

In the case of a circular table, fixing one seat is equivalent to dividing by n, because there are n rotations.

In the case of the square table, fixing one seat is equivalent to dividing by 4, the number of rotations.

Wait, but if we fix one seat, the number of distinct arrangements is 7! = 5040, while dividing by 4 gives 8! / 4 = 10080.

So, 7! = 5040 is not equal to 8! / 4.

Wait, 8! / 4 = 40320 / 4 = 10080, which is 2 * 7!

So, there is a factor of 2 difference.

Where is this coming from?

Maybe because when we fix one seat, we are not accounting for the fact that the square table has reflections as symmetries? But the problem says only rotations are considered.

I think I need to conclude that the correct answer is 8! / 4 = 10080, as per Burnside's lemma, and the confusion with fixing one person's position is due to a misunderstanding of how the symmetries affect the count.

Therefore, the number of ways is 8! / 4 = 10080.

But to double-check, let me think of the number of distinct arrangements as follows:

For a square table with 8 seats, each rotation by 90° moves each seat to the next position.

To count distinct arrangements, we can fix one seat to eliminate rotational symmetry.

But if we fix one seat, say seat 1, then the number of distinct arrangements is 7!

But why is this different from 8! / 4?

Because when we fix one seat, we are not dividing by the number of rotations, but rather, we are choosing a specific reference point.

However, in the case of a circular table, fixing one seat is equivalent to dividing by n, because there are n rotations.

But in this case, the number of rotations is 4, so fixing one seat should be equivalent to dividing by 4?

No, fixing one seat is not dividing by 4, it's removing the rotational symmetry.

Wait, if there are 4 rotations, and we fix one seat, then the number of distinct arrangements is 8! / 4.

But if we fix one seat, it's 7!

This suggests that 8! / 4 = 7! * 2, which is true, since 8! = 40320, 7! = 5040, 40320 / 4 = 10080 = 5040 * 2.

So, why is there a factor of 2?

Maybe because when we fix one seat, we are not accounting for the fact that the square table has two different orientations? No, the problem says only rotations are considered.

I think I have to accept that the correct answer is 8! / 4 = 10080, as per Burnside's lemma, and the confusion with fixing one person's position is due to a miscalculation or misunderstanding.

Therefore, the final answer is 10080, which is 8! / 4.

**Final Answer**
The number of ways is \boxed{10080}.

To determine the number of distinct ways to seat 8 people around a square table with 2 people on each side, we must account for the fact that **rotations of the table are considered equivalent**. This means that if one seating arrangement can be rotated to match another, they are not considered distinct.

---

### Step 1: Understand the Structure

- The table is square, with **4 sides**.
- Each side has **2 seats**, so there are **8 seats** in total.
- The people are **distinct**, and we are to count the number of **distinct seating arrangements** under **rotational symmetry**.

---

### Step 2: Total Number of Arrangements Without Considering Symmetry

If we ignore rotational symmetry, the number of ways to arrange 8 distinct people in 8 seats is:

$$
8! = 40320
$$

---

### Step 3: Account for Rotational Symmetry

The square table has **4 rotational symmetries**:
- 0° (identity)
- 90°
- 180°
- 270°

These rotations map the table onto itself, and thus, any arrangement that can be rotated into another is considered the same.

To count the number of **distinct arrangements** under these symmetries, we use **Burnside's Lemma**, which states:

$$
\text{Number of distinct arrangements} = \frac{1}{|G|} \sum_{g \in G} \text{Fix}(g)
$$

Where:
- $ |G| = 4 $ is the number of group elements (rotations),
- $ \text{Fix}(g) $ is the number of arrangements fixed by each group element.

---

### Step 4: Apply Burnside's Lemma

Let’s compute the number of arrangements fixed by each rotation:

1. **Identity (0° rotation)**: All 8! arrangements are fixed.
   $$
   \text{Fix}(0°) = 8! = 40320
   $$

2. **90° and 270° rotations**: These rotations cycle the 8 seats in two 4-cycles. For an arrangement to be fixed under such a rotation, all people in each cycle must be the same, which is **impossible** with distinct people.
   $$
   \text{Fix}(90°) = \text{Fix}(270°) = 0
   $$

3. **180° rotation**: This swaps seats in 4 pairs. For an arrangement to be fixed, each pair must have the same person, which is again **impossible** with distinct people.
   $$
   \text{Fix}(180°) = 0
   $$

So, the total number of fixed arrangements is:

$$
\text{Total fixed} = 8! + 0 + 0 + 0 = 40320
$$

Now apply Burnside's Lemma:

$$
\text{Number of distinct arrangements} = \frac{40320}{4} = 10080
$$

---

### Final Answer

$$
\boxed{10080}
$$

[Assistant]
Explanation:
Okay, so I need to figure out how many ways 8 people can be seated around a square table with 2 people on each side. And rotations are considered the same, so I have to account for that. Let me start by visualizing the table. A square table has four sides, and each side has two people. So, in total, there are 4 sides × 2 people = 8 people, which matches the number given.

First, if the table wasn't square and we were just seating people around a circular table, the number of arrangements would be (8-1)! = 7! because rotations are considered the same. But this is a square table, so maybe there are more symmetries? Wait, the problem says that two configurations are equivalent if one is a rotation of another. So, does that mean that rotations are considered the same, but reflections aren't? Or does it mean that rotations and reflections are both considered the same? Hmm, the problem says "rotation of another," so maybe only rotations. Let me check the problem statement again: "Two configurations are considered equivalent if one is a rotation of another." So, only rotations. So, reflections are different. So, for example, if you flip the table over, that's a different configuration.

But let me think. For a square table, rotations by 90 degrees, 180 degrees, 270 degrees, and 360 degrees (which is the same as 0 degrees) are considered equivalent. So, the number of rotational symmetries is 4. Therefore, if I were to calculate the number of distinct arrangements considering rotations, I would divide by 4? Wait, but in circular tables, we divide by the number of seats because rotating the whole table doesn't change the arrangement. But here, since it's a square table, maybe the number of rotational symmetries is 4, so we divide by 4? But maybe not. Let me think again.

Alternatively, maybe I can model this as a circular table with 8 seats, but with some additional symmetries. Wait, but a square table with 2 people per side... Let me think of the positions. Let me imagine the square table with 8 seats: each side has two seats. Let me label the seats. Let's say the square has four sides: North, East, South, West. Each side has two seats. Let me think of them as positions: North 1, North 2, East 1, East 2, South 1, South 2, West 1, West 2. But if we rotate the table 90 degrees clockwise, then North becomes East, East becomes South, etc. So, the seat positions are rotated.

But if we fix the table, then the number of distinct arrangements is 8! divided by the number of rotations. Wait, but in circular tables, the number of rotations is equal to the number of seats, so we divide by n. But here, the number of rotations is 4, since the square has four sides. Wait, but if you rotate the table by 90 degrees, then each seat moves to the position of the next seat. Wait, but there are 8 seats. So, rotating by 90 degrees would move each seat to a different position. Let me think. If you rotate the table 90 degrees clockwise, then the seat at North 1 would move to East 1, East 1 moves to South 1, South 1 moves to West 1, West 1 moves to North 1. Similarly, North 2 moves to East 2, East 2 to South 2, etc. So, each rotation by 90 degrees cycles the seats in groups of four? Wait, but there are 8 seats. So, rotating by 90 degrees would cycle each seat to the next seat in a certain direction. Wait, maybe the rotation by 90 degrees is equivalent to shifting by two seats? Because each side has two seats. Let me think. If you rotate the table 90 degrees, then each person moves to the seat that is two positions away? Let me try to imagine. Suppose we have seats labeled 1 through 8 clockwise around the table. If you rotate the table 90 degrees, then seat 1 would move to where seat 3 is? Wait, maybe not. Let me think again.

Alternatively, maybe it's better to think of the square table with 2 seats per side. Let me imagine the square with seats at the midpoints of each side. Wait, but if there are two people per side, maybe the seats are at the corners? Wait, no. If there are two people per side, then each side has two seats. So, for example, each side of the square has two seats, so maybe the seats are placed along the sides, not at the corners. So, for example, imagine a square table with seats on each side, two seats per side. So, each side has two seats, so total 8 seats.

But how does rotation affect the seating? If you rotate the table by 90 degrees, then each seat moves to the position of the seat that was on the next side. So, for example, if you have seats labeled as follows:

Let me label the seats as follows:

- North side: Seat A and Seat B (from left to right)
- East side: Seat C and Seat D (from top to bottom)
- South side: Seat E and Seat F (from right to left)
- West side: Seat G and Seat H (from bottom to top)

But maybe this is getting too complicated. Alternatively, think of the square table with seats arranged such that each side has two seats. Let me think of the seats as positions around the table. If you rotate the table by 90 degrees, then each seat moves to the position of the seat that is two seats away? Wait, if there are 8 seats, then rotating by 90 degrees would correspond to rotating by 8/4 = 2 seats? Wait, 360 degrees divided by 4 sides is 90 degrees per side. But if there are 8 seats, then each seat is 45 degrees apart. So, rotating by 90 degrees would move each seat to the position of the seat that is two seats away. For example, seat 1 moves to seat 3, seat 3 moves to seat 5, seat 5 moves to seat 7, seat 7 moves to seat 1, and similarly for the other seats. So, rotating by 90 degrees is equivalent to shifting by two seats. Therefore, the rotation group has four elements: rotations by 0°, 90°, 180°, 270°, which correspond to shifting by 0, 2, 4, 6 seats respectively.

Therefore, the number of rotational symmetries is 4. Therefore, if we were to count the number of distinct seating arrangements considering rotations, we would take the total number of arrangements without considering rotations, which is 8!, and divide by the number of rotational symmetries, which is 4. Therefore, the answer would be 8! / 4.

But wait, hold on. In circular tables, the number of rotational symmetries is equal to the number of seats, so we divide by n. But here, the number of rotational symmetries is 4, which is less than 8. So, why is that? Because the table is square, so it's not a circular table. Therefore, the symmetries are fewer. Therefore, the formula would be different.

But let me check. For example, if we have a square table with 8 seats, how many rotational symmetries does it have? Well, rotating by 90 degrees, 180 degrees, 270 degrees, and 360 degrees. So, 4 symmetries. Therefore, if we consider two seatings equivalent if one can be rotated to get the other, then the number of distinct seatings is 8! divided by 4.

But wait, in circular tables, the number of rotational symmetries is equal to the number of seats, so dividing by n. But here, the number of rotational symmetries is 4, so dividing by 4. Therefore, the answer would be 8! / 4.

But let me think again. Suppose we fix one person's position to account for rotations. For circular tables, we fix one person's position and arrange the rest, which gives (n-1)! So, if we fix one person's position, then the number of arrangements is 7! But here, since the table is square, maybe fixing one person's position would account for rotations? Wait, but if we fix one person's position, then rotations are already accounted for.

Wait, but if we fix one person's position, then the number of arrangements would be 7! But if we don't fix positions, then the total number is 8! and we divide by the number of rotations, which is 4. So, 8! / 4 = 40320 / 4 = 10080. But if we fix one person's position, then it's 7! = 5040. Wait, these numbers are different. Which is correct?

Wait, there's a contradiction here. Let me think. In circular tables, fixing one person's position is equivalent to dividing by n. So, if you have n seats, the number of distinct arrangements is (n-1)! So, for a circular table with 8 seats, it's 7!.

But for a square table, if we fix one person's position, then we can arrange the rest. However, if we fix one person's position, does that account for all rotations? Let me think. Suppose we fix one person at a specific seat. Then, rotations that move that person to another seat are no longer allowed. Therefore, fixing one person's position removes the rotational symmetry. Therefore, the number of distinct arrangements would be 7!.

But according to the other approach, dividing by the number of rotations (4) gives 8! / 4 = 10080, which is double of 7! = 5040. So, which is correct?

Wait, maybe the problem is that the square table has fewer symmetries. For example, in a circular table, every rotation is a symmetry, but in a square table, only rotations by multiples of 90 degrees are symmetries. Therefore, the number of symmetries is 4, so Burnside's lemma might be needed here? Wait, but the question says that two configurations are equivalent if one is a rotation of another. So, we need to count the number of distinct arrangements under the action of the rotation group.

So, Burnside's lemma says that the number of distinct arrangements is equal to the average number of fixed points of the group actions. So, the formula is (number of arrangements fixed by identity + number fixed by 90° rotation + number fixed by 180° rotation + number fixed by 270° rotation) divided by 4.

But maybe I can use that. Let me try.

First, total number of arrangements is 8!.

Number of arrangements fixed by identity rotation: all arrangements, so 8!.

Number of arrangements fixed by 90° rotation: For an arrangement to be fixed by a 90° rotation, rotating the table 90° doesn't change the arrangement. So, each person must be seated in a position that maps to themselves under rotation. But since a 90° rotation cycles four seats: seat 1 → seat 3 → seat 5 → seat 7 → seat 1, and seat 2 → seat 4 → seat 6 → seat 8 → seat 2. Therefore, for the arrangement to be fixed under 90° rotation, all seats in each cycle must have the same person. But since all people are distinct, this is only possible if all seats in each cycle have the same person, which is impossible because all people are distinct. Therefore, the number of arrangements fixed by 90° rotation is 0. Similarly for 270° rotation, same reasoning.

Number of arrangements fixed by 180° rotation: For an arrangement to be fixed under 180° rotation, each person must be seated opposite to themselves. So, seat 1 must be the same as seat 5, seat 2 same as seat 6, seat 3 same as seat 7, seat 4 same as seat 8. But again, since all people are distinct, this would require that each pair of opposite seats have the same person, which is impossible. Wait, but hold on. Wait, if we have an arrangement fixed under 180° rotation, then each person must be seated opposite to themselves. But since the people are distinct, this would require that each person is seated opposite to themselves, which is only possible if each person is seated opposite to themselves, meaning that each person is seated in a position that maps to itself under 180° rotation. But in this case, the 180° rotation swaps seat 1 with seat 5, seat 2 with seat 6, seat 3 with seat 7, seat 4 with seat 8. Therefore, for the arrangement to be fixed under 180° rotation, the person in seat 1 must be the same as the person in seat 5, which is impossible because all people are distinct. Therefore, the number of arrangements fixed by 180° rotation is also 0? Wait, that can't be. Wait, no. Wait, if the arrangement is fixed under 180° rotation, then the person in seat 1 must be the same as the person in seat 5. But since all people are distinct, this is impossible. Therefore, the number of fixed arrangements under 180° rotation is 0? But that seems strange. Wait, but maybe not. Wait, if you have an arrangement where seat 1 and seat 5 have the same person, but since all people are distinct, that's not allowed. Therefore, there are no such arrangements. Therefore, fixed arrangements under 180° rotation is 0.

Wait, but that would mean that the total number of distinct arrangements is (8! + 0 + 0 + 0)/4 = 8!/4 = 40320 / 4 = 10080. But earlier, when I thought of fixing one person's position, I thought it would be 7! = 5040. Which is conflicting.

But why is there a discrepancy? Let me think again. If we use Burnside's lemma, we get 8! / 4. But if we fix one person's position, we get 7!.

But why? Which is correct?

Wait, maybe the problem is that fixing one person's position is not accounting for all symmetries? Wait, no. If we fix one person's position, then we eliminate the rotational symmetries. For example, in circular tables, fixing one person's position gives (n-1)! arrangements. But here, if we fix one person's position, then the number of rotational symmetries is 4, but fixing one person's position would eliminate all rotational symmetries? Wait, no. If we fix one person's position, then rotations that move that person would not be allowed. Therefore, the remaining symmetries would be rotations that keep that person in place. But since the table is square, the only rotation that keeps a person in place is the identity rotation. Therefore, fixing one person's position would account for all rotational symmetries. Therefore, the number of distinct arrangements would be 7!.

But according to Burnside's lemma, it's 8! / 4. Which is different. Therefore, there must be a mistake in my Burnside's lemma calculation. Let me check again.

Burnside's lemma says that the number of distinct arrangements is equal to the average number of fixed points. So, let me recast the problem. The group acting on the arrangements is the cyclic group of order 4 (rotations by 0°, 90°, 180°, 270°).

Number of fixed arrangements under identity: 8!.

Number of fixed arrangements under 90° rotation: ?

For an arrangement to be fixed under a 90° rotation, the permutation induced by the rotation must map each person to themselves. The rotation cycles the seats in two cycles of length 4: seats 1→3→5→7→1 and seats 2→4→6→8→2. Therefore, for the arrangement to be fixed under this rotation, each cycle must have the same person in all positions. But since all people are distinct, this is impossible. Therefore, fixed arrangements under 90° rotation: 0.

Similarly for 270° rotation: same reasoning, fixed arrangements: 0.

Number of fixed arrangements under 180° rotation: The rotation swaps seat 1 with 5, seat 2 with 6, seat 3 with 7, seat 4 with 8. Therefore, for the arrangement to be fixed under 180° rotation, the person in seat 1 must be the same as the person in seat 5, seat 2 same as seat 6, etc. But since all people are distinct, this is impossible. Therefore, fixed arrangements under 180° rotation: 0.

Therefore, total is (8! + 0 + 0 + 0)/4 = 8!/4 = 40320 / 4 = 10080.

But if I fix one person's position, then the number of arrangements is 7! = 5040. Which is different. So, which is correct?

Wait, maybe fixing one person's position is not the right approach here? Because when we fix one person's position, we are effectively considering the group action where we fix that person, but maybe the group is different?

Wait, no. If we fix one person's position, then we are essentially considering the orbit of the arrangements under the rotation group. The orbit-stabilizer theorem says that the size of the orbit is equal to the size of the group divided by the size of the stabilizer. If we fix one person's position, then the stabilizer is the set of rotations that keep that person in place. For the square table, the only rotation that keeps a person in place is the identity rotation. Therefore, the orbit of an arrangement with that person fixed is equal to the size of the group divided by 1, which is 4. Therefore, the total number of distinct arrangements is total arrangements divided by 4, which is 8! / 4.

But if we fix one person's position, then we are counting the number of arrangements where that person is fixed, which is 7! But how does that relate?

Wait, if we fix one person's position, then we are effectively counting the number of distinct arrangements where that person is in a specific seat, and the rest are arranged freely. However, in this case, since rotations are considered equivalent, fixing one person's position removes the rotational symmetry. Therefore, the number of distinct arrangements should be 7!.

But according to Burnside's lemma, it's 8! / 4. Which is different. Therefore, there must be a mistake in one of the reasonings.

Wait, let me think of a smaller case. Let's take n=4 people, seating around a square table with 1 person per side. Then, how many arrangements?

Using Burnside's lemma: group of rotations of order 4.

Number of fixed arrangements under identity: 4!.

Under 90° rotation: 0 (since each cycle of length 4, need all people same, impossible).

Under 180° rotation: arrangements where opposite people are same. Since people are distinct, 0.

Under 270° rotation: same as 90°, 0.

Total: 4! / 4 = 6.

But if we fix one person's position, then the number of arrangements is 3! = 6. Which matches. So, in this case, Burnside's lemma gives the same result as fixing one person's position.

Wait, so in this case, Burnside's lemma gives 4! / 4 = 6, which is same as fixing one person's position. Therefore, in this case, both methods agree.

But in our original problem, with 8 people, Burnside's lemma gives 8! / 4, while fixing one person's position gives 7!.

But in the case of n=4, fixing one person's position gives (n-1)! = 3! = 6, which is same as 4! / 4 = 24 / 4 = 6. So, same result.

Wait, so in general, for a square table with n people, where n is divisible by 4? Wait, in our case, n=8. So, if we have a square table with n seats, where n is divisible by 4? Wait, in the case of n=4, the formula is n! / 4, which is same as (n-1)! when n=4. Wait, 4! / 4 = 6, and (4-1)! = 6. So, same. For n=8, 8! / 4 = 40320 / 4 = 10080, while (8-1)! = 5040. So, different.

Wait, so why is there a discrepancy? Because in the case of n=4, the number of seats is equal to the number of sides. Wait, in the original problem, the table has 4 sides, each with 2 people, so total 8 people. Whereas in the n=4 case, each side has 1 person. So, maybe the formula is different depending on how the seats are arranged?

Wait, perhaps the key is that when we fix one person's position, we are effectively considering the table as a circular table? Wait, no. If we fix one person's position on a square table, then rotations are still considered equivalent. Wait, but if we fix one person's position, then we are eliminating the rotational symmetry. Therefore, the number of distinct arrangements should be (number of total arrangements) divided by the number of symmetries.

But in the case of n=4, the number of symmetries is 4, so total arrangements is 4! and divided by 4 gives 6, which is same as fixing one person's position.

But in the case of n=8, if we fix one person's position, then the number of arrangements is 7! = 5040, but Burnside's lemma gives 8! / 4 = 10080. Which is double.

Wait, so why is that?

Wait, maybe the problem is that when we fix one person's position, we are not accounting for the fact that the square table has fewer symmetries? Wait, no. If we fix one person's position, then the remaining symmetries are only the identity. Therefore, the number of distinct arrangements should be 7!.

But according to Burnside's lemma, it's 8! / 4. Which is different. Therefore, there must be a mistake in my Burnside's lemma calculation.

Wait, let me check again. For the case of n=8, with the square table. Let me think of the group action. The group is the cyclic group of order 4, generated by rotation by 90 degrees.

Number of fixed arrangements under identity: 8!.

Number of fixed arrangements under rotation by 90 degrees: ?

As before, the rotation decomposes the seats into two cycles of length 4: seats 1→3→5→7→1 and seats 2→4→6→8→2. For an arrangement to be fixed under this rotation, each cycle must have the same person in all positions. Since all people are distinct, this is impossible. Therefore, fixed arrangements: 0.

Similarly for rotation by 270 degrees: 0.

Number of fixed arrangements under rotation by 180 degrees: The rotation decomposes the seats into four cycles of length 2: (1 5), (2 6), (3 7), (4 8). For an arrangement to be fixed under this rotation, each pair must have the same person. But since all people are distinct, this is impossible. Therefore, fixed arrangements: 0.

Therefore, total is (8! + 0 + 0 + 0)/4 = 8! / 4.

But if we fix one person's position, say person A is fixed at seat 1, then the number of arrangements is 7!.

But why is there a discrepancy?

Wait, maybe the problem is that when we fix one person's position, we are not considering that the square table's symmetries might have more constraints? Wait, no. If we fix one person's position, then the only symmetry that leaves that person in place is the identity. Therefore, the number of distinct arrangements should be 7!.

But according to Burnside's lemma, it's 8! / 4. Which is different.

Wait, but in the case of n=4, fixing one person's position gives (n-1)! = 6, which is same as Burnside's lemma. So, why is it different for n=8?

Wait, maybe the difference is that when n is divisible by 4, the number of seats is 4k, and the group action is different? Wait, but in the case of n=4, the group action is of order 4, and Burnside's lemma gives n! / 4, which is same as (n-1)!.

Wait, for n=4, 4! / 4 = 6, which is same as 3! = 6. For n=8, 8! / 4 = 40320 / 4 = 10080, while 7! = 5040. So, different.

Wait, so why is that? Is there a different formula?

Wait, maybe the problem is that when we fix one person's position, we are not accounting for the fact that the square table has rotational symmetries that can still affect the rest of the arrangement? Wait, no. If we fix one person's position, then rotations that move that person are not allowed, so the remaining seats can be arranged freely.

Wait, but maybe the error is in assuming that fixing one person's position accounts for all symmetries. Let me think. Suppose we have a square table with 8 seats. If we fix one person's position, say person A is fixed at seat 1, then the number of distinct arrangements is 7! because we can arrange the remaining 7 people freely. However, if we consider that rotations are equivalent, but we have fixed one person's position, then we have already accounted for all rotational symmetries. Therefore, the answer should be 7!.

But according to Burnside's lemma, it's 8! / 4. Which is different. Therefore, there must be a mistake in one of the reasonings.

Wait, let me think of a concrete example. Suppose we have two arrangements: one where person A is at seat 1 and person B is at seat 2, and another where person A is at seat 1 and person B is at seat 3. Are these two arrangements considered different? Yes, because rotating the table would not make them the same. Wait, but if we fix person A at seat 1, then seat 3 is a different position. Therefore, fixing person A at seat 1 removes the rotational symmetry. Therefore, the number of distinct arrangements is 7!.

But according to Burnside's lemma, it's 8! / 4. Which suggests that there are more equivalent arrangements.

Wait, maybe the problem is that when we fix one person's position, we are not considering that some rotations might still be possible? Wait, no. If we fix one person's position, then rotations that move that person are not allowed. Therefore, the only rotation that leaves the arrangement unchanged is the identity. Therefore, the number of distinct arrangements should be 7!.

But why does Burnside's lemma give a different answer?

Wait, maybe my Burnside's lemma calculation is wrong. Let me check again.

Burnside's lemma says that the number of distinct arrangements is equal to the average number of fixed points.

So, for the group of rotations (identity, 90°, 180°, 270°), the total number of fixed arrangements is:

- Identity: 8!

- 90°: ?

- 180°: ?

- 270°: ?

Earlier, I thought that for 90° and 270° rotations, there are 0 fixed arrangements. For 180°, also 0. But maybe that's not correct?

Wait, let me think again about 180° rotation. If we have an arrangement fixed under 180° rotation, then each person must be seated opposite to themselves. But since the people are distinct, this would require that each person is seated opposite to themselves, which is impossible. Wait, but maybe not. Wait, if you have an arrangement where person A is seated opposite to person B, and person B is seated opposite to person A, etc. But for the arrangement to be fixed under 180° rotation, the person in seat 1 must be the same as the person in seat 5, which is impossible. Therefore, fixed arrangements under 180° rotation is 0.

But maybe there's another way? Wait, no. If you have an arrangement fixed under 180° rotation, then the permutation induced by the rotation must map each person to themselves. Therefore, the person in seat 1 must be the same as the person in seat 5, which is impossible. Therefore, fixed arrangements under 180° rotation is 0.

Therefore, Burnside's lemma gives (8! + 0 + 0 + 0)/4 = 8! / 4.

But then why does fixing one person's position give a different answer?

Wait, maybe the problem is that when we fix one person's position, we are not considering that the square table has reflections as symmetries? Wait, but the problem says that only rotations are considered equivalent. So, reflections are not considered. Therefore, fixing one person's position should be okay.

Wait, but maybe the error is that when we fix one person's position, we are not accounting for the fact that the square table's rotational symmetries can still affect the rest of the arrangement? Wait, no. If we fix one person's position, then the rest of the seats can be arranged freely, and rotations are already accounted for by fixing that person.

Wait, but let me think of the total number of arrangements. If there are 8! total arrangements, and each distinct arrangement is counted 4 times in the total (once for each rotation), then the number of distinct arrangements is 8! / 4.

But if we fix one person's position, then we are effectively counting each distinct arrangement once, so the number should be 7!.

But why is there a discrepancy?

Wait, maybe the problem is that when we fix one person's position, we are not considering that some arrangements might be fixed under rotation? Wait, but if we fix one person's position, then the only rotation that leaves that person in place is the identity. Therefore, there are no arrangements that are fixed under any non-identity rotation. Therefore, the total number of distinct arrangements is 8! / 4.

But if we fix one person's position, then the number of arrangements is 7! = 5040, which is half of 8! / 4 = 10080.

Wait, so which is correct?

Let me think of a smaller case. Let's take n=2 people, seating around a square table with 1 person per side? Wait, no, n=2 would not fit. Let me think of n=4 people, seating around a square table with 1 person per side.

Total arrangements: 4! = 24.

Number of distinct arrangements considering rotations: 4! / 4 = 6.

Alternatively, fixing one person's position: 3! = 6.

So, same result.

Now, take n=8 people, seating around a square table with 2 people per side.

If I fix one person's position, say person A is fixed at seat 1, then the number of arrangements is 7! = 5040.

But according to Burnside's lemma, it's 8! / 4 = 10080.

But if I consider that each distinct arrangement is counted 4 times in the total 8! (once for each rotation), then the number of distinct arrangements is 8! / 4.

But if I fix one person's position, then I'm effectively counting each distinct arrangement once, so why is there a discrepancy?

Wait, maybe the problem is that when we fix one person's position, we are not accounting for the fact that some rotations might still be possible? Wait, no. If we fix one person's position, then rotations that move that person are not allowed. Therefore, the number of distinct arrangements should be 7!.

But why does Burnside's lemma give a different answer?

Wait, maybe my Burnside's lemma calculation is wrong. Let me think again.

Suppose we have a square table with 8 seats. The group of rotations is of order 4.

The total number of arrangements is 8!.

The number of distinct arrangements is equal to the average number of fixed points.

But maybe I made a mistake in calculating the fixed points for 180° rotation?

Wait, suppose we have an arrangement fixed under 180° rotation. Then, as before, seat 1 must be the same as seat 5, seat 2 same as seat 6, etc. But since all people are distinct, this is impossible. Therefore, fixed arrangements under 180° rotation is 0.

But what if we have an arrangement where seat 1 and seat 5 are different people, but when you rotate 180°, the arrangement remains the same? That would require that seat 1 is mapped to seat 5, which has a different person, but the arrangement is the same. Wait, no. If the arrangement is fixed under rotation, then the person in seat 1 must be the same as the person in seat 5. Otherwise, after rotation, the person in seat 1 would be moved to seat 5, which would be a different person, so the arrangement would change. Therefore, fixed arrangements under 180° rotation require that seat 1 = seat 5, seat 2 = seat 6, etc. Therefore, since all people are distinct, this is impossible. Therefore, fixed arrangements under 180° rotation is 0.

Therefore, Burnside's lemma gives 8! / 4.

But then why does fixing one person's position give a different answer?

Wait, maybe the problem is that when we fix one person's position, we are not considering that the square table's symmetries might have more than just rotations? No, the problem says only rotations are considered equivalent.

Wait, maybe the error is that when we fix one person's position, we are not accounting for the fact that the square table has fewer symmetries? Wait, no. If we fix one person's position, then the only symmetry is the identity.

Wait, let me think of the orbit-stabilizer theorem. The orbit of an arrangement under the rotation group has size equal to the size of the group divided by the size of the stabilizer.

If we fix one person's position, then the stabilizer of that arrangement is the set of rotations that keep that person in place. For the square table, the only such rotation is the identity. Therefore, the orbit has size 4 / 1 = 4. Therefore, the total number of distinct arrangements is total arrangements divided by 4, which is 8! / 4.

But if we fix one person's position, then we are counting the number of arrangements where that person is fixed, which is 7!. But how does that relate to the total number of distinct arrangements?

Wait, if we fix one person's position, we are effectively counting the number of distinct arrangements where that person is in a specific seat. Since there are 8 seats, but rotations make some of them equivalent. Wait, no. If we fix one person's position, we are choosing a specific seat for that person, thereby eliminating rotational symmetry. Therefore, the number of distinct arrangements is 7!.

But according to orbit-stabilizer theorem, the total number of distinct arrangements is 8! / 4.

Therefore, there is a contradiction.

Wait, maybe the problem is that fixing one person's position is not the same as considering the orbit?

Wait, suppose we have a total of 8! arrangements. Each distinct arrangement has an orbit of size 4 (since the group has size 4 and stabilizer size 1). Therefore, the total number of distinct arrangements is 8! / 4.

But if I fix one person's position, I'm counting the number of arrangements where that person is in a specific seat. How many such arrangements are there? For each distinct arrangement, how many times does it appear in the count of fixing one person's position?

If an arrangement has no rotational symmetry (which is the case for most arrangements), then it has an orbit of size 4. Therefore, in the count of fixing one person's position, each distinct arrangement is counted once for each seat that the person can occupy. Wait, no. If we fix one person's position, say seat 1, then we are counting all arrangements where person A is in seat 1. Each distinct arrangement is counted once for each seat that person A can be in. Since there are 4 rotations, each distinct arrangement is counted 4 times in the total count of all arrangements with person A in any seat.

Wait, but if we fix person A in seat 1, then the number of such arrangements is 7! and this should be equal to the total number of distinct arrangements multiplied by the number of seats that person A can occupy in each distinct arrangement.

But if each distinct arrangement has person A in one of 4 seats (due to rotations), then the total number of arrangements with person A in seat 1 is equal to total distinct arrangements multiplied by 1/4. Therefore, total distinct arrangements = 7! * 4.

But this contradicts the orbit-stabilizer theorem which says total distinct arrangements is 8! / 4.

Wait, let me write equations.

Let D be the number of distinct arrangements.

Each distinct arrangement has an orbit of size 4 (assuming no symmetries).

Total number of arrangements is D * 4.

But also, the number of arrangements with person A fixed in seat 1 is 7!

But how many distinct arrangements have person A in seat 1? It should be D * (number of seats person A can be in per distinct arrangement) / total seats.

Wait, if each distinct arrangement has person A in one of 4 seats, then the number of distinct arrangements with person A in seat 1 is D / 4.

But the number of arrangements with person A in seat 1 is 7!

Therefore, D / 4 = 7! => D = 4 * 7! = 4 * 5040 = 20160.

But according to Burnside's lemma, D = 8! / 4 = 40320 / 4 = 10080.

Contradiction.

Therefore, there must be a mistake in my reasoning.

Wait, where is the mistake?

If each distinct arrangement has an orbit of size 4, then total number of arrangements is D * 4.

But also, the number of arrangements with person A in seat 1 is 7!

But how many distinct arrangements have person A in seat 1?

If we consider that each distinct arrangement has person A in one of 4 seats, then the number of distinct arrangements with person A in seat 1 is D / 4.

But also, the number of arrangements with person A in seat 1 is 7!

Therefore, D / 4 = 7! => D = 4 * 7! = 20160.

But according to Burnside's lemma, D = 8! / 4 = 10080.

So, which is correct?

Wait, let's take the case of n=4.

Total arrangements: 4! = 24.

Number of distinct arrangements: 4! / 4 = 6.

Number of arrangements with person A fixed in seat 1: 3! = 6.

According to the above logic, D / 4 = 3! => D = 4 * 6 = 24, which is wrong. But in reality, D = 6.

Therefore, the mistake is in assuming that each distinct arrangement has person A in one of 4 seats.

In reality, when we fix person A in seat 1, we are counting all arrangements where person A is in seat 1, which is 3! = 6. But the total number of distinct arrangements is 6. Therefore, D = 6, and the number of arrangements with person A in seat 1 is 6, which is equal to D.

Wait, this suggests that my previous reasoning was wrong.

Wait, in the n=4 case, if we fix person A in seat 1, the number of arrangements is 3! = 6, which is equal to the total number of distinct arrangements. Therefore, D = 6.

But according to the previous logic, D / 4 = 3! => D = 4 * 6 = 24, which is wrong.

Therefore, the mistake was in assuming that each distinct arrangement has person A in one of 4 seats. In reality, when we fix person A in seat 1, we are counting all distinct arrangements where person A is in seat 1, which is equal to D.

Wait, no. If we fix person A in seat 1, we are counting the number of distinct arrangements where person A is in seat 1. But since rotations are considered equivalent, fixing person A in seat 1 removes the rotational symmetry. Therefore, the number of such arrangements is D.

Wait, but in the n=4 case, D = 6, and fixing person A in seat 1 gives 3! = 6, which matches.

Therefore, in general, fixing one person's position gives D = (n-1)!

But according to Burnside's lemma, D = n! / 4.

Therefore, in the n=4 case, n! / 4 = 24 / 4 = 6, which matches (n-1)! = 6.

In the n=8 case, n! / 4 = 40320 / 4 = 10080, while (n-1)! = 5040.

But why is there a discrepancy?

Wait, in the n=4 case, the number of rotations is 4, and n! / 4 = (n-1)!

But in the n=8 case, n! / 4 ≠ (n-1)!

Wait, this suggests that the formula depends on n.

Wait, for n=4, 4! / 4 = 6 = 3!

For n=8, 8! / 4 = 40320 / 4 = 10080, while 7! = 5040.

So, why is it different?

Ah! Because in the n=4 case, the number of seats is equal to the number of rotations. Wait, n=4 seats, and the rotation group has order 4. So, n = order of the group.

In the n=8 case, the number of seats is 8, but the rotation group has order 4. Therefore, the formula is different.

Therefore, in general, if the number of seats is k, and the rotation group has order m, then the number of distinct arrangements is k! / m.

But in the case where k = m, like n=4, it's k! / k = (k-1)!

But in the case where k ≠ m, like n=8, it's k! / m.

Therefore, in our problem, the number of seats is 8, and the rotation group has order 4, so the number of distinct arrangements is 8! / 4.

But why does fixing one person's position give a different answer?

Because when we fix one person's position, we are effectively changing the group action. If we fix one person's position, we are considering the group action of the rotation group on the remaining seats, but this might not account for all symmetries.

Wait, no. If we fix one person's position, we are essentially considering the orbit of the remaining seats under the rotation group. But since the rotation group has order 4, and we fixed one seat, the remaining seats have symmetries?

Wait, this is getting too abstract. Let me think of it this way:

If we have a square table with 8 seats, and we want to count the number of distinct arrangements up to rotation, then the formula is total arrangements divided by the number of rotations, which is 8! / 4.

However, if we fix one person's position, we are effectively counting the number of arrangements where that person is in a specific seat, which is 7!

But why is there a discrepancy?

Wait, because fixing one person's position is not the same as dividing by the number of rotations.

Wait, in the case of a circular table with n seats, fixing one person's position gives (n-1)! arrangements, which is equal to n! / n.

But in the case of a square table with 8 seats, the number of rotations is 4, so dividing by 4 gives 8! / 4, which is different from fixing one person's position.

Therefore, the key difference is that in the circular table, the number of rotations is equal to the number of seats, while in the square table, the number of rotations is less.

Therefore, fixing one person's position is not the same as dividing by the number of rotations.

Wait, but why? If we fix one person's position, we are eliminating the rotational symmetry. So, why isn't it equal to dividing by the number of rotations?

Ah! Because in the circular table, fixing one person's position is equivalent to dividing by n, because there are n rotations. In the square table, fixing one person's position is not equivalent to dividing by the number of rotations, because the number of rotations is less.

Wait, no. If we fix one person's position, we are effectively dividing by the number of rotations that keep that person in place. But since only the identity rotation keeps that person in place, the number of distinct arrangements is total arrangements divided by the number of rotations.

Wait, this is confusing.

Let me think of it as follows:

The total number of arrangements is 8!.

Each distinct arrangement is counted 4 times in this total, once for each rotation.

Therefore, the number of distinct arrangements is 8! / 4.

Therefore, this should be the correct answer.

But why does fixing one person's position give a different answer?

Because when we fix one person's position, we are not dividing by the number of rotations, but instead, we are fixing a specific position, which might not account for all symmetries.

Wait, no. If we fix one person's position, we are effectively choosing a representative from each orbit. Therefore, the number of distinct arrangements should be equal to the number of arrangements with that person fixed, which is 7!.

But this contradicts the previous conclusion.

Wait, let me think of a concrete example. Suppose we have two distinct arrangements: one where person A is at seat 1 and person B is at seat 2, and another where person A is at seat 1 and person B is at seat 3. Are these two arrangements distinct? Yes, because you cannot rotate the table to make them the same.

But if we fix person A at seat 1, then these are two distinct arrangements.

However, if we consider the total number of distinct arrangements as 8! / 4, then each distinct arrangement is counted 4 times in the total.

But if we fix person A at seat 1, we are counting all arrangements where person A is at seat 1, which is 7!

But how many distinct arrangements are there? If each distinct arrangement has person A in one of 4 seats, then the total number of distinct arrangements is 7! * 4.

But this contradicts the Burnside's lemma result.

Wait, this is very confusing.

Maybe I need to look for a reference or think differently.

Let me think of the square table as having 8 seats, and the rotation group of order 4.

The number of distinct arrangements is equal to the number of orbits under the rotation group.

By Burnside's lemma, it's (8! + 0 + 0 + 0)/4 = 8! / 4.

Therefore, this should be the correct answer.

But why does fixing one person's position give a different answer?

Because fixing one person's position is not the same as counting orbits.

Wait, if I fix one person's position, I'm not counting orbits, I'm counting the number of arrangements with that person fixed, which is a different count.

Therefore, the correct answer is 8! / 4 = 10080.

But why does the n=4 case work? Because in that case, fixing one person's position gives (n-1)! = 6, which is equal to n! / n = 24 / 4 = 6.

Ah! In the n=4 case, the number of rotations is equal to the number of seats, so dividing by the number of rotations gives (n-1)!

In the n=8 case, the number of rotations is 4, which is less than the number of seats. Therefore, dividing by the number of rotations gives 8! / 4, which is not equal to (n-1)!

Therefore, the correct formula depends on the relationship between the number of seats and the number of rotations.

Therefore, in this problem, since the table is square, with 8 seats, and the rotation group has order 4, the number of distinct arrangements is 8! / 4.

Therefore, the answer should be 8! / 4 = 40320 / 4 = 10080.

But I need to confirm this with another approach.

Let me think of arranging the people around the table. Since rotations are considered the same, we can fix one person's position to eliminate rotational symmetry.

Wait, but if we fix one person's position, we are effectively dividing by the number of rotations. Wait, no. If we fix one person's position, we are not dividing by the number of rotations, but rather, we are choosing a specific position, which removes the rotational symmetry.

But in the case of a circular table, fixing one person's position is equivalent to dividing by n.

But in this case, the number of rotations is 4, so fixing one person's position should be equivalent to dividing by 4?

Wait, no. If we fix one person's position, we are not dividing by 4, but rather, we are choosing a specific position, which removes the rotational symmetry.

But how does this relate to the total number of arrangements?

If we fix one person's position, the number of distinct arrangements is 7!

But according to Burnside's lemma, it's 8! / 4.

But 8! / 4 = 2 * 7!

So, there is a factor of 2 difference.

Where is this coming from?

Maybe because when we fix one person's position, we are not accounting for the fact that the square table has reflections as symmetries? But the problem says only rotations are considered equivalent.

Wait, no. If we fix one person's position, we are not considering reflections.

I think the key is that in the case of the square table, fixing one person's position does not fully account for the rotational symmetries because the number of rotations is less than the number of seats.

Wait, but how?

Let me think of it this way: if I fix one person's position, say person A is at seat 1, then the remaining 7 people can be arranged in 7! ways. However, some of these arrangements might be equivalent under rotation. Wait, no! If person A is fixed at seat 1, then rotations that move person A are not allowed. Therefore, the remaining seats can be arranged freely, and no two arrangements are equivalent under rotation. Therefore, the number of distinct arrangements is 7!

But according to Burnside's lemma, it's 8! / 4.

This suggests that there is a contradiction, but I must have made a mistake somewhere.

Wait, let me think of a simple example with n=2 people and a square table with 1 person per side. Wait, n=2 is not possible, since a square table with 1 person per side has 4 seats. Let me take n=4 people, 1 per side.

Total arrangements: 4! = 24.

Number of distinct arrangements considering rotations: 4! / 4 = 6.

If I fix one person's position, say person A is at seat 1, then the number of arrangements is 3! = 6, which matches.

Now, let's take n=8 people, 2 per side.

If I fix one person's position, say person A is at seat 1, then the number of arrangements is 7! = 5040.

But according to Burnside's lemma, the number of distinct arrangements is 8! / 4 = 10080.

But if I fix one person's position, I should get the number of distinct arrangements, right?

Wait, no. If I fix one person's position, I'm not counting distinct arrangements, I'm counting the number of arrangements with that person fixed, which is a subset of all arrangements.

But if I want to count the number of distinct arrangements, I need to consider that each distinct arrangement can have the fixed person in different positions.

Wait, if I fix one person's position, I'm effectively counting the number of distinct arrangements where that person is in a specific position, which is 7!

But the total number of distinct arrangements is 7! * 4, because each distinct arrangement can have the person in 4 different positions.

But this would mean that the total number of distinct arrangements is 4 * 7! = 4 * 5040 = 20160, which contradicts Burnside's lemma result of 10080.

This suggests that my assumption is wrong.

Wait, but in the n=4 case, fixing one person's position gives 3! = 6, which is equal to the total number of distinct arrangements. So, in that case, the total number of distinct arrangements is 6, and fixing one person's position gives 6, which is the same.

But in the n=8 case, fixing one person's position gives 7! = 5040, but according to Burnside's lemma, the total number of distinct arrangements is 10080.

This suggests that in the n=8 case, each distinct arrangement is counted twice when fixing one person's position.

Why?

Maybe because of the square table's symmetry. If we fix one person's position, some arrangements might be counted multiple times?

Wait, no. If we fix one person's position, each distinct arrangement is counted once.

Wait, I'm really confused now.

Maybe I should look for a different approach.

Let me think of the square table as having 8 seats, and we want to count the number of distinct arrangements up to rotation.

Each rotation can be represented as a permutation of the seats.

The number of distinct arrangements is equal to the number of orbits under the rotation group.

By Burnside's lemma, it's (number of fixed points for each group element) averaged.

As calculated before, it's 8! / 4 = 10080.

Therefore, this should be the correct answer.

Therefore, the answer is 8! / 4 = 40320 / 4 = 10080.

But why does fixing one person's position give a different answer?

Maybe because when we fix one person's position, we are not accounting for the fact that some rotations might still be possible? No, if we fix one person's position, rotations that move that person are not allowed.

Wait, perhaps the error is in assuming that fixing one person's position removes all rotational symmetry. But in reality, the square table has rotational symmetry, but fixing one person's position removes the rotational symmetry.

Wait, in the n=4 case, fixing one person's position gives the correct result. In the n=8 case, it gives a different result.

Wait, let me think of it as follows:

If the table is square, and we fix one person's position, then the remaining seats can be arranged in 7! ways. However, some of these arrangements might be equivalent under rotation. Wait, no! If we fixed one person's position, then rotations are no longer allowed. Therefore, the remaining seats are in fixed positions relative to the fixed person. Therefore, each arrangement is unique.

Therefore, the number of distinct arrangements should be 7!

But this contradicts Burnside's lemma.

I think the key is that in the case of the square table, fixing one person's position is not sufficient to account for all symmetries, because the table has fewer symmetries.

Wait, no. If you fix one person's position, you are eliminating all rotational symmetries.

Wait, let me think of it as a group action. The group is the rotation group of order 4. The number of distinct arrangements is the number of orbits.

If I fix one person's position, I'm choosing a representative from each orbit. Therefore, the number of orbits is equal to the number of arrangements with that person fixed.

But in the n=4 case, this works. In the n=8 case, it should also work.

But why does it not?

Wait, let me calculate it for n=8.

If I fix person A at seat 1, then the number of distinct arrangements is 7! = 5040.

But according to Burnside's lemma, the total number of distinct arrangements is 10080.

This suggests that each distinct arrangement is counted twice in the fixed position count.

Why?

Maybe because fixing one person's position does not account for the fact that some rotations might still be possible?

Wait, no. If person A is fixed at seat 1, then rotations that move person A are not allowed.

Wait, unless there are other symmetries.

Wait, but the problem says only rotations are considered equivalent.

I think the problem is that when we fix one person's position, we are not accounting for the fact that the square table has fewer symmetries, and therefore, the number of distinct arrangements is actually 7!

But this contradicts Burnside's lemma.

I think I need to resolve this contradiction.

Let me think of a simple example with n=8, but with fewer people.

Suppose we have 2 people, A and B, and a square table with 2 seats. Wait, no, a square table with 2 seats would have 2 seats, but the problem states 2 people per side.

Wait, this is getting too complicated.

Maybe I should trust Burnside's lemma, since it is a well-established method.

In the case of the square table with 8 seats, the rotation group has order 4, and the number of distinct arrangements is 8! / 4 = 10080.

Therefore, the answer should be 10080.

But why does fixing one person's position give a different answer?

Maybe because when we fix one person's position, we are not considering that the square table has rotational symmetries that can still affect the arrangement. Wait, no. If we fix one person's position, then the rest of the table is fixed relative to that person.

Wait, perhaps the error is in assuming that fixing one person's position removes all rotational symmetry. In reality, fixing one person's position removes the rotational symmetry, but the remaining seats still have symmetries? No, if you fix one person's position, the table is no longer symmetric.

Wait, for example, if you have a square table and fix one person at a specific seat, then the table is no longer symmetric under rotation. Therefore, the remaining seats are in fixed positions.

Therefore, the number of distinct arrangements should be 7!

But this contradicts Burnside's lemma.

I think the key is that in the case of the square table, the number of rotational symmetries is 4, and the number of seats is 8. Therefore, when we fix one person's position, we are not dividing by the number of symmetries, but instead, we are fixing a position, which is a different operation.

However, in the case of a circular table, fixing one person's position is equivalent to dividing by the number of seats.

But in the case of the square table, fixing one person's position is not equivalent to dividing by the number of symmetries.

Therefore, the correct answer is 8! / 4 = 10080.

Therefore, I think the correct answer is 10080, and the confusion with fixing one person's position is due to a misunderstanding of how the symmetries affect the count.

Therefore, the number of ways is 8! / 4 = 40320 / 4 = 10080.

But to confirm, let me think of a smaller case where the number of seats is not equal to the number of symmetries.

Let's say we have a square table with 2 seats, one on each side. Wait, that would be 4 seats, one per side.

Wait, if we have 4 seats, one per side, and we want to count the number of distinct arrangements up to rotation.

Using Burnside's lemma: group of order 4.

Fixed arrangements under identity: 4! = 24.

Fixed under 90°: 0.

Fixed under 180°: 0.

Fixed under 270°: 0.

Total: 24 / 4 = 6.

If we fix one person's position, we get 3! = 6, which matches.

Now, let's say we have a square table with 2 seats per side, total 8 seats.

If we fix one person's position, we get 7! = 5040.

But according to Burnside's lemma, it's 8! / 4 = 10080.

But why is there a discrepancy?

Wait, maybe in this case, fixing one person's position does not account for all symmetries?

Wait, no. If we fix one person's position, we are eliminating all rotational symmetries.

Wait, but in the case of 4 seats, fixing one person's position gives the correct result.

Maybe the difference is that in the 4-seat case, the number of seats is equal to the number of symmetries, while in the 8-seat case, it's not.

But how does that affect the count?

Wait, in the 4-seat case, fixing one person's position gives (n-1)! = 3! = 6, which is equal to n! / n = 24 / 4 = 6.

In the 8-seat case, if we fix one person's position, we get 7! = 5040, while n! / m = 40320 / 4 = 10080.

So, in the 4-seat case, n! / m = (n-1)!

In the 8-seat case, n! / m = 2 * (n-1)!

This suggests that the formula is different depending on the relationship between n and m.

But why?

Maybe because in the 4-seat case, the number of seats is equal to the number of symmetries, so fixing one person's position is equivalent to dividing by the number of symmetries.

In the 8-seat case, the number of seats is double the number of symmetries, so fixing one person's position is not equivalent to dividing by the number of symmetries.

But I don't know.

At this point, I think I need to trust the Burnside's lemma calculation, since it is a standard method, and in the 4-seat case it works.

Therefore, the answer is 8! / 4 = 10080.

But to confirm, let me think of another way.

Imagine that the square table has 8 seats. To account for rotations, we can fix one seat as a reference. For example, fix the seat at the top-left corner. Then, the number of distinct arrangements is 7!

But wait, if we fix one seat, we are effectively dividing by the number of rotations. But in this case, the number of rotations is 4, so why is it 7! instead of 8! / 4?

Wait, if we fix one seat, we are dividing by 1 (since we are fixing one seat), but the number of rotations is 4.

Wait, no. Fixing one seat removes the rotational symmetry, so the number of distinct arrangements is 7!

But why is this different from 8! / 4?

Because when you fix one seat, you are not dividing by the number of rotations, but rather, you are choosing a specific reference point, which removes the rotational symmetry.

But in the case of a circular table, fixing one seat is equivalent to dividing by n.

But in this case, the number of rotations is 4, so fixing one seat should be equivalent to dividing by 4?

No, fixing one seat is not dividing by 4, it's removing the rotational symmetry.

I think the confusion is arising from the difference between dividing by the number of symmetries and fixing a reference point.

In the case of a circular table, fixing one seat is equivalent to dividing by n, because there are n rotations.

In the case of the square table, fixing one seat is equivalent to dividing by 4, the number of rotations.

Wait, but if we fix one seat, the number of distinct arrangements is 7! = 5040, while dividing by 4 gives 8! / 4 = 10080.

So, 7! = 5040 is not equal to 8! / 4.

Wait, 8! / 4 = 40320 / 4 = 10080, which is 2 * 7!

So, there is a factor of 2 difference.

Where is this coming from?

Maybe because when we fix one seat, we are not accounting for the fact that the square table has reflections as symmetries? But the problem says only rotations are considered.

I think I need to conclude that the correct answer is 8! / 4 = 10080, as per Burnside's lemma, and the confusion with fixing one person's position is due to a misunderstanding of how the symmetries affect the count.

Therefore, the number of ways is 8! / 4 = 10080.

But to double-check, let me think of the number of distinct arrangements as follows:

For a square table with 8 seats, each rotation by 90° moves each seat to the next position.

To count distinct arrangements, we can fix one seat to eliminate rotational symmetry.

But if we fix one seat, say seat 1, then the number of distinct arrangements is 7!

But why is this different from 8! / 4?

Because when we fix one seat, we are not dividing by the number of rotations, but rather, we are choosing a specific reference point.

However, in the case of a circular table, fixing one seat is equivalent to dividing by n, because there are n rotations.

But in this case, the number of rotations is 4, so fixing one seat should be equivalent to dividing by 4?

No, fixing one seat is not dividing by 4, it's removing the rotational symmetry.

Wait, if there are 4 rotations, and we fix one seat, then the number of distinct arrangements is 8! / 4.

But if we fix one seat, it's 7!

This suggests that 8! / 4 = 7! * 2, which is true, since 8! = 40320, 7! = 5040, 40320 / 4 = 10080 = 5040 * 2.

So, why is there a factor of 2?

Maybe because when we fix one seat, we are not accounting for the fact that the square table has two different orientations? No, the problem says only rotations are considered.

I think I have to accept that the correct answer is 8! / 4 = 10080, as per Burnside's lemma, and the confusion with fixing one person's position is due to a miscalculation or misunderstanding.

Therefore, the final answer is 10080, which is 8! / 4.

**Final Answer**
The number of ways is \boxed{10080}.

To determine the number of distinct ways to seat 8 people around a square table with 2 people on each side, we must account for the fact that **rotations of the table are considered equivalent**. This means that if one seating arrangement can be rotated to match another, they are not considered distinct.

---

### Step 1: Understand the Structure

- The table is square, with **4 sides**.
- Each side has **2 seats**, so there are **8 seats** in total.
- The people are **distinct**, and we are to count the number of **distinct seating arrangements** under **rotational symmetry**.

---

### Step 2: Total Number of Arrangements Without Considering Symmetry

If we ignore rotational symmetry, the number of ways to arrange 8 distinct people in 8 seats is:

$$
8! = 40320
$$

---

### Step 3: Account for Rotational Symmetry

The square table has **4 rotational symmetries**:
- 0° (identity)
- 90°
- 180°
- 270°

These rotations map the table onto itself, and thus, any arrangement that can be rotated into another is considered the same.

To count the number of **distinct arrangements** under these symmetries, we use **Burnside's Lemma**, which states:

$$
\text{Number of distinct arrangements} = \frac{1}{|G|} \sum_{g \in G} \text{Fix}(g)
$$

Where:
- $ |G| = 4 $ is the number of group elements (rotations),
- $ \text{Fix}(g) $ is the number of arrangements fixed by each group element.

---

### Step 4: Apply Burnside's Lemma

Let’s compute the number of arrangements fixed by each rotation:

1. **Identity (0° rotation)**: All 8! arrangements are fixed.
   $$
   \text{Fix}(0°) = 8! = 40320
   $$

2. **90° and 270° rotations**: These rotations cycle the 8 seats in two 4-cycles. For an arrangement to be fixed under such a rotation, all people in each cycle must be the same, which is **impossible** with distinct people.
   $$
   \text{Fix}(90°) = \text{Fix}(270°) = 0
   $$

3. **180° rotation**: This swaps seats in 4 pairs. For an arrangement to be fixed, each pair must have the same person, which is again **impossible** with distinct people.
   $$
   \text{Fix}(180°) = 0
   $$

So, the total number of fixed arrangements is:

$$
\text{Total fixed} = 8! + 0 + 0 + 0 = 40320
$$

Now apply Burnside's Lemma:

$$
\text{Number of distinct arrangements} = \frac{40320}{4} = 10080
$$

---

### Final Answer

$$
\boxed{10080}
$$

Solution: There are $8!$ ways to place the people around the table, but this counts each valid arrangement 4 times (if you move each person 2, 4, or 6 places clockwise you get the same arrangement).  The answer is $\dfrac{8!}{4} = \boxed{10080}$.