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 configuration. Hmm, let me think.

First, I remember that arranging people around a table usually involves circular permutations. But this is a square table, so maybe there's a difference? For a circular table, the number of arrangements is (n-1)! because rotations are considered the same. But a square table, even though it's a polygon, might have different symmetries? Wait, but the problem says that rotations are considered equivalent. So maybe it's similar to a circular table but with some restrictions?

Wait, a square table has four sides. If there are two people on each side, that means each side has two seats? So total seats would be 4 sides * 2 people = 8 people. Yep, that matches the question. So, it's like a square with seats on each side, two per side.

But how does that affect the number of arrangements? For circular tables, we fix one person's position to account for rotations. But here, since it's a square, maybe rotations by 90 degrees would be considered equivalent? Wait, the problem says "rotations of another", so maybe rotations by any multiple of 90 degrees? Let me check.

If we rotate the table 90 degrees, the positions of the people would shift. But the problem states that two configurations are equivalent if one is a rotation of another. So, for example, if you rotate the table by 90 degrees, the arrangement is considered the same. Therefore, similar to circular permutations, but the number of rotational symmetries is different. For a square, there are four rotational symmetries: 0°, 90°, 180°, and 270°. So, the number of rotational symmetries is 4. Therefore, maybe the formula is similar to circular permutations but divided by 4 instead of n?

Wait, for a circular table with n seats, the number of distinct arrangements is (n-1)! because there are n rotations that result in equivalent configurations. But here, since there are only 4 rotational symmetries, maybe the formula is (number of linear arrangements)/4? Let me think.

Alternatively, maybe I should fix one person's position to account for rotations. If I fix one person at a specific seat, then arrange the others. For example, fix one person at a specific seat, then arrange the remaining 7 people. But wait, if the table is square, fixing one person might fix the rotational symmetry?

Wait, but if we fix one person at a particular seat, then rotations are already accounted for. For example, in a circular table, fixing one person removes the rotational symmetry. Similarly, for the square table, if I fix one person at a seat, then the rest can be arranged relative to that person. However, the square also has reflections, but the problem doesn't mention reflections being equivalent. It only mentions rotations. So reflections would be considered different.

So maybe the formula is similar to circular permutations but divided by 4? Let me check with a smaller number. Suppose there are 4 people, one on each side of a square table. How many arrangements? If we consider rotations equivalent, then it's (4-1)! = 6? Wait, but if there are four seats, and rotations are considered equivalent, the number is (4-1)! = 6. But if we fix one person, then we have 3! = 6. So that works. However, if we consider the square's rotational symmetry, there are 4 rotations. So in general, for n seats around a table with rotational symmetry of order k, the formula is n!/k? Wait, for circular table, n seats, rotational symmetry of order n, so number of arrangements is n!/n = (n-1)!.

But here, with the square table, if there are 8 seats, and rotational symmetry of order 4, then the number of arrangements would be 8!/4? But is that correct?

Wait, let me think again. Suppose we have a square table with 8 seats: two on each side. Let me imagine the seats as follows: seat 1 and seat 2 on the north side, seat 3 and seat 4 on the east side, seat 5 and seat 6 on the south side, seat 7 and seat 8 on the west side. Then, rotating the table 90 degrees clockwise would move seat 1 to seat 3's position, seat 2 to seat 4, seat 3 to seat 5, seat 4 to seat 6, seat 5 to seat 7, seat 6 to seat 8, seat 7 to seat 1, seat 8 to seat 2. So, each rotation moves each seat to the next side's position.

So if we consider two seatings equivalent if one can be rotated to get the other, then how many distinct arrangements are there?

In general, the formula for counting distinct arrangements under rotational symmetry is given by Burnside's lemma, which averages the number of fixed points of the group actions. But maybe that's overcomplicating? Wait, but if I can just divide by the number of symmetries?

Wait, but Burnside's lemma is more reliable here. Let me recall. Burnside's lemma says that the number of distinct arrangements is equal to the average number of elements fixed by each group action. So, for the group of rotations, which has 4 elements (0°, 90°, 180°, 270°), we need to compute the number of arrangements fixed by each rotation, then add them up and divide by 4.

Alternatively, if all rotations except the identity have no fixed arrangements, then the total number would be (total arrangements + 0 + 0 + 0)/4 = total arrangements /4. But is that the case?

Wait, let me check. For the identity rotation (0°), all arrangements are fixed, so that's 8!.

For the 90° rotation, how many arrangements are fixed? That would mean that rotating the table 90° leaves the arrangement unchanged. So, each person must be seated in a position that maps to themselves when rotated. But since the rotation cycles all 8 seats in a single cycle? Wait, no. Let me think.

If you rotate the table 90°, then each seat moves to the next seat on the adjacent side. For example, seat 1 goes to seat 3, seat 3 goes to seat 5, seat 5 goes to seat 7, seat 7 goes to seat 1. Similarly, seat 2 goes to seat 4, seat 4 goes to seat 6, seat 6 goes to seat 8, seat 8 goes to seat 2. Therefore, the rotation decomposes the seats into two cycles of length 4: one cycle for the odd-numbered seats (1,3,5,7) and one cycle for the even-numbered seats (2,4,6,8). For an arrangement to be fixed under a 90° rotation, the people in each cycle must be the same. But since all people are distinct, this is only possible if all people in each cycle are the same. But since all people are distinct, this is impossible. Therefore, there are 0 fixed arrangements under 90° rotation.

Similarly, for the 270° rotation, same logic applies. It's equivalent to rotating -90°, which also cycles the seats in the same manner, so same result: 0 fixed arrangements.

Now for the 180° rotation. This rotation swaps seat 1 with seat 5, seat 2 with seat 6, seat 3 with seat 7, seat 4 with seat 8. So, each person must be seated opposite their original position. Therefore, for an arrangement to be fixed under 180° rotation, each person must be seated opposite to themselves, meaning that each pair (1 and 5, 2 and 6, etc.) must have the same person. But again, since all people are distinct, this is only possible if each pair has the same person, which is impossible. Wait, hold on, no. Wait, if you have person A at seat 1, then person A must also be at seat 5. But since all people are distinct, you can't have two people at the same seat. Therefore, actually, there are no fixed arrangements under 180° rotation as well? Wait, that can't be right. Wait, maybe I made a mistake here.

Wait, no. Wait, the arrangement is fixed under 180° rotation if rotating the table 180° results in the same arrangement. That would mean that person at seat 1 is the same as person at seat 5, etc. But since seats are occupied by different people, this would require that person at seat 1 is the same as person at seat 5, which can't happen unless they are the same person. But since all people are distinct, there are no such arrangements. Therefore, fixed arrangements under 180° rotation would be zero? Wait, but that would mean that Burnside's lemma gives us (8! + 0 + 0 + 0)/4 = 8!/4. But is this correct?

But that seems conflicting with my intuition. Wait, for example, with n=4 seats at a square table, with one seat per side. Let's do that case. If there are 4 seats, one on each side. How many arrangements considering rotations equivalent?

By Burnside's lemma: group has 4 rotations. Identity fixes 4! = 24. Rotations by 90° and 270° fix 0 arrangements (since each cycle is length 4, requires all seats to have same person). Rotation by 180°: fixes arrangements where opposite seats are same. So how many? For 4 seats, opposite seats must be same. But since people are distinct, we need to pair them. Wait, but if you have 4 people, and you need to assign them to seats such that seat 1 = seat 3 and seat 2 = seat 4. So the number of such arrangements is 4! / (2! * 2!) ? Wait, no. Wait, actually, you have to assign people to pairs. So, first choose a person for seat 1 and 3, then a person for seat 2 and 4. So, it's 4 choices for seat 1, then 3 choices for seat 2, and then the rest are determined. But since seat 1 and 3 are the same, but they are different seats. Wait, actually, the number of arrangements fixed under 180° rotation would be the number of ways to assign people such that seat 1 and seat 3 are same, seat 2 and seat 4 are same. But since people are distinct, you can't have two people in the same seat. Therefore, actually, there are zero such arrangements? Wait, that can't be. Wait, maybe I need to think differently.

Wait, if the arrangement is fixed under 180° rotation, then person in seat 1 must be the same as person in seat 3. But since each seat is occupied by a different person, this is impossible. Therefore, there are zero arrangements fixed under 180° rotation. Therefore, Burnside's lemma gives (24 + 0 + 0 + 0)/4 = 6. Which is equal to (4-1)! = 6. So that works. Therefore, for n=4, this formula works. So maybe in the original problem with 8 seats, the Burnside's lemma gives 8! / 4. Which would be 40320 / 4 = 10080. But that seems high?

Wait, but let's check with another example. Suppose n=2 seats on a square table? Wait, that's not possible, since square has four sides. Maybe n=2 people? Hmm, not sure. Maybe n=8 is okay.

But wait, if we fix one person, then arrange the rest, how does that work?

If we fix one person at a seat, then the remaining 7 can be arranged in 7! ways. But since rotations are considered equivalent, fixing one person accounts for rotational symmetry. However, in the square case, fixing one person would eliminate rotational symmetry? Wait, for example, suppose I fix person A at seat 1. Then, rotations that move seat 1 to another seat are eliminated. Therefore, the number of distinct arrangements would be 7!.

But earlier Burnside's lemma suggested 8! / 4 = 10080, which is 8!/4, and 7! is 5040. Which is different. So which is correct?

Wait, there's a contradiction here. So which one is correct?

Wait, let me think again. For circular table with n seats, fixing one person gives (n-1)! arrangements. For the square table with 8 seats, if we fix one person, we are effectively removing rotational symmetry? But the square has rotational symmetry of order 4, so fixing one person might not fully account for all symmetries? Wait, no. If you fix one person at a specific seat, then rotations that move that seat are eliminated. So for example, if you fix person A at seat 1, then the other rotations would move seat 1 to other positions, but since we've fixed person A there, those rotations are not allowed. Therefore, the number of distinct arrangements is 7!.

But according to Burnside's lemma above, it was 8!/4 = 10080, which is 8!/4 = 40320 / 4 = 10080, whereas 7! is 5040. These are different. So which is correct?

Wait, there must be an error in one of the approaches. Let me think again about Burnside's lemma. For the square table with 8 seats, the group of rotations is of order 4: rotations by 0°, 90°, 180°, 270°. Let me recalculate the fixed arrangements.

For identity rotation: all 8! arrangements are fixed.

For rotation by 90°: an arrangement is fixed if rotating 90° leaves it unchanged. As discussed earlier, this would require that each person is mapped to themselves under rotation. However, the rotation decomposes the seats into two cycles of length 4: one cycle for the seats on the "odd" positions (1,3,5,7) and another cycle for the "even" positions (2,4,6,8). For the arrangement to be fixed, the person in seat 1 must be the same as seat 3, seat 3 same as seat 5, etc. Since all people are distinct, this requires that all these seats have the same person, which is impossible. Therefore, fixed arrangements under 90° rotation: 0.

Similarly for 270° rotation: same result, 0.

For rotation by 180°: each seat is mapped to its opposite seat. So seat 1 swaps with 5, 2 swaps with 6, 3 swaps with 7, 4 swaps with 8. Therefore, for the arrangement to be fixed, the person in seat 1 must be the same as seat 5, person in seat 2 same as seat 6, etc. Again, since all people are distinct, this is impossible. Therefore, fixed arrangements under 180° rotation: 0.

Therefore, total fixed arrangements: 8! + 0 + 0 + 0 = 40320. Divided by 4, gives 40320 / 4 = 10080. But if I fix one person, I get 7! = 5040. Which one is correct?

Wait, if we fix one person, we need to consider that fixing a person removes the rotational symmetry. However, in the case of the square table, maybe fixing one person is not sufficient? Wait, no. If you fix person A at seat 1, then rotations that move seat 1 to other positions are no longer allowed. Therefore, the number of distinct arrangements should be 7!.

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

Wait, let's take a small example again. Suppose n=4 seats, one per side. Then, according to Burnside's lemma, it would be (4! + 0 + 0 + 0)/4 = 24 / 4 = 6. Which is (4-1)! = 6. So that works. Now, if we fix one person, say seat 1, then we have 3! = 6, which is same as Burnside's result. So in that case, both methods agree.

Wait, so why in the case of n=8 seats, fixing one person gives 7! but Burnside's lemma gives 8! / 4? Because 7! = 5040 and 8! / 4 = 10080. They are different. So that suggests that there is an error in one of the reasonings.

Wait, let me think again about Burnside's lemma for n=8 seats. If I fix one person, then I can think of it as dividing by the number of rotations. But in the case where the number of rotations is 4, then dividing by 4 would be the same as fixing one position?

Wait, but in circular table, we have n rotations. So to fix one position, divide by n. Here, for square table, number of rotations is 4, so divide by 4. But in the case of n=4 seats, dividing by 4 gives (4-1)! = 6. Which is same as fixing one person. So maybe here, with 8 seats, fixing one person is equivalent to dividing by 4? But how?

Wait, if I fix one person, then the number of arrangements is 7! = 5040. According to Burnside's lemma, it's 8!/4 = 40320 / 4 = 10080. Which is double. Therefore, something is wrong here.

Wait, perhaps I made a mistake in Burnside's lemma calculation. Let me check again.

Wait, when we fix a person, we are effectively considering arrangements up to rotation. But perhaps in the case of the square table, fixing one person doesn't fully account for the symmetries? Wait, no. Fixing one person should eliminate all rotational symmetries.

Alternatively, maybe the Burnside's lemma was applied incorrectly. Let me re-examine.

Burnside's lemma says the number of distinct arrangements is equal to (number of arrangements fixed by each group element averaged over the group). So group has 4 elements: identity, rotation 90°, 180°, rotation 270°.

Number fixed by identity: 8!.

Number fixed by 90°: as before, 0.

Number fixed by 180°: let's re-examine.

If we have an arrangement fixed under 180° rotation, that means that each person is seated opposite to themselves. Wait, no. It means that when you rotate the table 180°, the arrangement looks the same. Therefore, the person at seat 1 must be the same as the person at seat 5. But since all people are distinct, this can't happen. Therefore, number of fixed arrangements under 180° rotation is 0? But that can't be, because maybe there are arrangements where people are arranged in pairs?

Wait, no. If the person at seat 1 is person A, then after rotation, person A must be at seat 5. But seat 5 is a different seat. Therefore, unless person A is sitting at both seat 1 and seat 5, which is impossible, there can be no fixed arrangements. Hence, fixed arrangements under 180° rotation is 0. Then Burnside's lemma gives (8! + 0 + 0 + 0)/4 = 8!/4. However, if we fix one person, we get 7!.

This discrepancy suggests that one of the methods is wrong. Wait, maybe fixing a person doesn't account for the entire symmetry?

Wait, take n=4. With Burnside's lemma, we get (4! + 0 + 0 + 0)/4 = 6. Which is same as fixing one person (3! = 6). So in that case, it works. What about with n=8 seats?

Wait, suppose we fix one person. Let me think of the square table with 8 seats. If we fix person A at seat 1, then the remaining 7 seats can be filled in 7! ways. However, are there still rotational symmetries left? For example, rotating the table 180°, which would swap seat 1 with seat 5. But seat 1 is fixed with person A, so seat 5 must also have person A. But that would require person A to be in two seats, which is impossible. Therefore, fixing person A eliminates all rotational symmetries? Because any rotation would require some person to be in multiple seats, which is impossible. Therefore, fixing one person should give the correct number of arrangements, which is 7!.

But according to Burnside's lemma, it's 8! / 4. Which is different. So there must be an error in Burnside's calculation. Wait, let's check Burnside's lemma again for n=8 seats.

Wait, perhaps my analysis of fixed arrangements under 180° is wrong. Let me think again. If we have an arrangement fixed under 180° rotation, that means that person at seat i is the same as person at seat i + 4 (modulo 8). However, since all seats are distinct, this would require that each pair of opposite seats has the same person. But since people are distinct, this is only possible if each pair is assigned the same person, which is impossible unless you have duplicate people. Therefore, there are zero such arrangements. Hence, Burnside's lemma calculation is correct?

But why is there a discrepancy?

Wait, maybe the group action isn't just rotations? Wait, the problem says "rotations of another", so only rotations. So group is cyclic group of order 4. But perhaps my mistake is in considering the seats as labeled? Wait, if the seats are unlabeled, then Burnside's lemma applies. But if seats are labeled, then the count is different. Wait, in our case, are the seats labeled? Or are they considered unlabeled?

Wait, when we talk about seating arrangements, the seats are considered distinct because they are different positions around the table. However, when considering rotations, we consider two arrangements equivalent if one can be rotated to get the other. So in Burnside's lemma, we need to count colorings (arrangements) up to rotation. So if we have 8 seats, each seat can be assigned a distinct person. So it's like coloring the seats with 8 different colors, considering rotations equivalent.

So Burnside's lemma says the number is (fixed by identity + fixed by 90 + fixed by 180 + fixed by 270)/4.

As above, fixed by identity is 8!.

Fixed by 90 and 270: 0.

Fixed by 180: ?

Wait, maybe my previous conclusion that fixed by 180 is zero is wrong. Wait, suppose we have an arrangement fixed under 180° rotation. That means that the person at seat 1 is same as seat 5, seat 2 same as seat 6, etc. However, since each person is unique, this would require that person at seat 1 equals person at seat 5, which is impossible. Therefore, fixed arrangements under 180° is zero. Therefore, Burnside's lemma gives 8! / 4.

But then why does fixing one person give a different answer? There must be something wrong here.

Wait, let's think of fixing person A at seat 1. Then, the rest can be arranged freely. However, in this case, are we accounting for all rotations? Suppose we don't fix person A, but consider rotations. So the total number of arrangements is 8!. Each equivalence class under rotation has size equal to...?

In general, for a group action, the number of equivalence classes is equal to the total number of arrangements divided by the size of the group, if all orbits have the same size. However, this is only true if no non-identity element fixes any arrangement. Which is the case here, because we saw that only identity fixes arrangements. Therefore, the number of equivalence classes is 8! / 4. Hence, Burnside's lemma gives 8! / 4. Therefore, fixing one person must not be the correct approach here?

Wait, but how? If I fix one person, then I should be able to count the number of equivalence classes. Wait, fixing one person is equivalent to considering orbits where one seat is fixed. Wait, maybe not. Let me think.

If we fix one person, then we are essentially choosing a representative from each orbit by fixing that person in that seat. But the number of orbits would be equal to the number of such arrangements. However, if fixing one person gives 7! arrangements, then the total number of orbits should be 7! = 5040, which is different from Burnside's count of 8! / 4 = 10080. Therefore, clearly, one of the logic is wrong.

Wait, here's a key point. When you fix one person, you are effectively counting the number of distinct arrangements where that person is in a specific seat. However, in the Burnside's lemma, we count all equivalence classes, where each equivalence class can have different numbers of arrangements. Wait, but if all orbits have size equal to the group size, which is 4, then the number of orbits is 8! / 4. However, if there are some orbits with smaller size, then Burnside's lemma counts them properly.

But earlier, when we considered fixed arrangements under non-identity rotations, we found that there are none. Therefore, all orbits have size equal to the group size, which is 4. Therefore, number of orbits is 8! / 4. Therefore, fixing one person should not give you the same number. Wait, but how?

Wait, the number of orbits is equal to the number of distinct arrangements under rotation. If I fix one person, then how does that relate?

Suppose I fix person A at seat 1. Then, the number of arrangements where person A is at seat 1 is 7!. However, each orbit contains exactly 4 different arrangements (rotated versions). However, some of these orbits may contain arrangements where person A is at seat 1, seat 3, seat 5, or seat 7. Wait, no. If an orbit contains an arrangement where person A is at seat 1, then rotating it by 90°, 180°, 270°, would move person A to seat 3, seat 5, seat 7. Therefore, each orbit that contains an arrangement with person A at seat 1 also contains arrangements with person A at seats 3, 5, 7. Therefore, the number of orbits where person A is at seat 1 is equal to the total number of orbits divided by 4 (since each orbit has 4 positions where person A can be). Therefore, total number of orbits is 4 * (number of arrangements with person A at seat 1) = 4 * 7! = 4 * 5040 = 20160. But according to Burnside's lemma, total number of orbits is 8! / 4 = 40320 / 4 = 10080. Which is half of 20160. So that suggests inconsistency.

Therefore, there is a contradiction here. Where is the mistake?

Wait, maybe the orbits have size 4, but the number of orbits is 8! / 4 = 10080. Therefore, the number of orbits where person A is at seat 1 is 10080 / 4 = 2520. But according to fixing person A at seat 1, we have 7! = 5040. So 2520 vs 5040. So there is a factor of 2 difference. Therefore, something is wrong here.

Wait, perhaps orbits are not size 4. Maybe some orbits have smaller size? But earlier we said that no non-identity element fixes any arrangement, so by orbit-stabilizer theorem, the size of each orbit is equal to the size of the group divided by the size of the stabilizer subgroup. Since the stabilizer is trivial (only identity), the size of each orbit is 4. Therefore, if there are 10080 orbits, each of size 4, the total number of arrangements is 10080 * 4 = 40320 = 8!, which is correct. Therefore, the number of orbits where person A is at seat 1 is equal to total number of orbits multiplied by the probability that a random arrangement has person A at seat 1. Since there are 8 seats, the probability is 1/8. Therefore, the number of such orbits is 10080 * (1/8) = 1260. But according to fixing person A at seat 1, we have 7! = 5040. Which is not equal to 1260. Therefore, clearly something is wrong here.

Wait, perhaps I have misunderstood the orbit-stabilizer theorem? Let me recall. Orbit-stabilizer theorem says that the size of the orbit of an element is equal to the size of the group divided by the size of the stabilizer. In our case, the stabilizer of an arrangement is the set of rotations that leave it unchanged. Since we have no non-identity rotations that stabilize any arrangement, the stabilizer is trivial, so the orbit has size 4. Therefore, the number of orbits is 8! / 4 = 10080. However, if we fix person A at seat 1, we are counting the number of arrangements with person A at seat 1, which is 7!.

But how does that relate to the number of orbits?

Each orbit contains 4 arrangements, each with person A at different seats. If we consider all orbits, each orbit contains exactly one arrangement with person A at seat 1, seat 3, seat 5, seat 7. Wait, but if person A is at seat 1, then after rotation, they can be at seat 3, 5, 7. So for each orbit that contains an arrangement with person A at seat 1, there are 4 different arrangements in the orbit with person A at different seats. Therefore, the number of such orbits is equal to the number of arrangements with person A at seat 1 divided by 4. Wait, but if we have 7! arrangements with person A at seat 1, then the number of orbits containing such arrangements is 7! / 4? But how?

Wait, no. Each orbit has exactly 4 arrangements. If an orbit contains an arrangement with person A at seat 1, then it contains 4 such arrangements, each with person A at different seats. Therefore, the number of orbits that contain an arrangement with person A at seat 1 is equal to the number of arrangements with person A at seat 1 divided by 4, but since each orbit is counted once for each seat that person A can be in. Wait, this is getting too convoluted.

Wait, let's think of it this way. Total number of arrangements with person A at seat 1 is 7!. Each such arrangement is in an orbit of size 4. However, in each orbit, there are 4 arrangements, each with person A at seat 1, seat 3, seat 5, seat 7. Therefore, the number of orbits that contain an arrangement with person A at seat 1 is equal to the total number of arrangements with person A at seat 1 divided by 4. Wait, no. Each orbit has exactly one arrangement with person A at seat 1, one at seat 3, etc. So the total number of orbits is equal to the number of arrangements with person A at seat 1 multiplied by 4? No, that would overcount.

Wait, I think I need to use the orbit-stabilizer theorem again. The total number of arrangements with person A at seat 1 is 7!. Each orbit that contains such an arrangement has size 4. However, each orbit contains exactly one arrangement where person A is at seat 1, one at seat 3, one at seat 5, one at seat 7. Therefore, the number of orbits that contain an arrangement with person A at seat 1 is equal to the number of such arrangements divided by 4? No, it's equal to the number of such arrangements divided by the number of arrangements per orbit. Wait, each orbit contains exactly one arrangement with person A at seat 1. Wait, no. If you have an orbit, and person A is in one of the seats, then rotating the arrangement would move person A to different seats. So if an orbit contains an arrangement with person A at seat 1, it must also contain arrangements with person A at seat 3, seat 5, seat 7. Therefore, each orbit that contains person A at any seat contains 4 arrangements with person A at different seats. Therefore, the number of orbits that contain person A at seat 1 is equal to the number of orbits that contain person A in any seat, divided by 4. But the total number of arrangements with person A at any seat is 8 * 7! / 8 = 7! (wait, no). Wait, the total number of arrangements where person A is at any seat is 8 * 7! / 8 = 7!? No, total number of arrangements is 8!. The number of arrangements where person A is at seat 1 is 7!; same for any seat. Therefore, total number of arrangements where person A is at any seat is 8 * 7! / 8 = 7!? No, that doesn't make sense. Wait, total number of arrangements where person A is at some specific seat is 7!. Since there are 8 seats, the total number of arrangements where person A is at any seat is 8 * 7! / 8? No, that's not. Actually, the total number is 8 * 7! but this counts each arrangement multiple times. Wait, no. Each arrangement has person A at exactly one seat. Therefore, total number of arrangements where person A is at any seat is just equal to the total number of arrangements, which is 8!. But this counts each arrangement once. So, the number of arrangements where person A is at seat 1 is 7!, and same for other seats.

Therefore, the total number of arrangements where person A is at seat 1 is 7!. Each such arrangement is in an orbit of size 4. How many orbits are there? Each orbit contains 4 arrangements. But how many of them have person A at seat 1?

If there are N orbits, then the number of arrangements where person A is at seat 1 is equal to N multiplied by the number of arrangements in each orbit that have person A at seat 1. Since each orbit contains exactly one arrangement with person A at seat 1 (wait, no). If you have an orbit, and person A is in one of the seats, then rotating it would place person A in 4 different seats. So each orbit contains exactly 4 arrangements where person A is in different seats. Therefore, if we count all arrangements where person A is at seat 1, it is equal to N multiplied by (number of orbits containing such an arrangement) * 1 (since each orbit contributes one such arrangement). Wait, this is confusing.

Alternatively, if total number of arrangements with person A at seat 1 is 7!, and each orbit contains exactly 4 such arrangements (one for each position), then total number of orbits is 7! / 4. But total number of orbits is also equal to 8! / 4 according to Burnside's lemma. Therefore:

7! / 4 = 8! / 4 → 7! = 8! → 7! = 8 * 7! → 1 = 8, which is false. Therefore, contradiction. Hence, my assumption is wrong.

Therefore, my previous assumption that each orbit contains exactly 4 arrangements with person A in different seats is wrong.

Wait, maybe not. Suppose I have an orbit containing an arrangement where person A is at seat 1. Then, rotating it by 90°, person A moves to seat 3, by 180° to seat 5, by 270° to seat 7. Hence, in this orbit, there are four arrangements: person A at seat 1, 3, 5, 7. Therefore, the number of orbits that contain an arrangement with person A at seat 1 is equal to the number of such arrangements divided by 4. But the number of such arrangements is 7! (arrangements with person A at seat 1). Therefore, the number of orbits is 7! / 4. But according to Burnside's lemma, it should be 8! / 4 = 2 * 7!.

So, where is the problem? It seems that there are two different results. Therefore, one of the reasonings must be wrong.

Wait, the problem is that when I fix person A at seat 1, I am considering arrangements where person A is fixed, and then others are permuted. However, in Burnside's lemma, we count all arrangements up to rotation. So, if I fix person A at seat 1, I am effectively choosing a representative from each orbit where person A is at seat 1. However, the total number of orbits is equal to the number of such representatives times the number of seats person A can be in per orbit. But if each orbit contains person A in four different seats, then number of orbits is (number of representatives) * 4. Wait, no. If I have M orbits, each containing 4 arrangements where person A is in different seats, then the number of arrangements with person A at seat 1 is M * 1 (since each orbit contributes one arrangement where person A is at seat 1). Therefore, M = number of arrangements with person A at seat 1 = 7! Therefore, number of orbits is 7! But this contradicts Burnside's lemma result of 8! / 4. So I need to resolve this.

Let me use concrete numbers with n=4. For n=4 seats (one per side), Burnside's lemma says 4! / 4 = 6. If I fix person A at seat 1, then number of arrangements is 3! = 6. Which matches. But according to the previous logic, if I fix person A at seat 1, number of orbits is 3! = 6, which matches Burnside's result. So why in the case of n=8, it doesn't match?

Wait, let me try n=2. Wait, n=2? Not sure. Maybe n=8 is a different case.

Wait, let me think again. For n=4, fixing person A at seat 1 gives number of arrangements 3! = 6. Burnside gives 4! / 4 = 6. Same result. For n=8, if we fix person A at seat 1, we get 7! = 5040. If Burnside's lemma gives 8! / 4 = 10080. Why the discrepancy?

Wait, perhaps when n is divisible by the number of rotations? In the case of n=4, the number of seats is equal to the number of rotations? No, n=4 seats, rotations of order 4. For n=8 seats, rotations of order 4. So why the difference?

Wait, let's try applying Burnside's lemma for n=4, but with the same logic. For n=4 seats, rotations by 90°, 180°, 270°, identity.

Fixed by identity: 4!

Fixed by 90° and 270°: 0.

Fixed by 180°: number of arrangements fixed under 180° rotation. For n=4, this would be arrangements where person at seat 1 equals seat 3 and seat 2 equals seat 4. Since people are distinct, this is 0? But wait, no! Wait, for n=4, if you fix under 180° rotation, you need the arrangement to be symmetric under 180°. So person at seat 1 must be same as seat 3, and person at seat 2 same as seat 4. But since all people are distinct, this is impossible. Hence, fixed arrangements under 180° is 0. Then Burnside gives (24 + 0 + 0 + 0)/4 = 6, which equals (4-1)! = 6. But if we fix one person, we get 3! = 6. So it works.

But for n=8, Burnside says 8! / 4, but fixing person gives 7! = 5040. Let's compute 8! / 4 = 40320 / 4 = 10080. And 7! = 5040. So why is there a factor of 2 difference?

Wait, maybe there's an error in assuming that all orbits have size 4. But for n=4, all orbits have size 4, and it works. For n=8, same reasoning: if no non-identity rotation fixes any arrangement, then all orbits have size 4, so number of orbits is 8! / 4. But why does fixing one person give a different answer?

Wait, maybe fixing one person doesn't account for some symmetries? Or maybe the table has more symmetries?

Wait, the problem says "Two configurations are considered equivalent if one is a rotation of another." So it's only rotations, not reflections. But what about reflections? If we consider reflections as different, then they are not part of the group. So group is only rotations.

Wait, but if we fix one person, how does that relate?

Wait, let me try with n=4 seats, fixing one person. Suppose seat 1 is fixed with person A. Then the remaining 3 people can be arranged in 3! ways. This corresponds to the orbits where person A is at seat 1. Each orbit has 4 arrangements, so total number of orbits is 3! * 4 / 4 = 3! ? No, it's total number of orbits is 3! ? Wait, no. If there are 3! arrangements with person A at seat 1, and each orbit contains 4 arrangements, then the number of orbits is 3! / 4? But that would be 6 / 4, which is not an integer. Contradiction. Wait, this suggests something is wrong.

But in reality, for n=4, we know the number of orbits is 6. If we fix person A at seat 1, we have 3! = 6 arrangements. Each such arrangement is in an orbit of size 4? But that would mean total number of arrangements is 6 * 4 = 24, which is correct. But each orbit is counted 4 times, one for each position of person A. So total number of orbits is 6 / 4 * 4? No, I'm confused again.

Wait, here's the key. Each orbit contains exactly one arrangement with person A at seat 1, one with person A at seat 2, etc. But in reality, for n=4, each orbit contains 4 arrangements, each with person A at different seats. Therefore, the number of orbits is equal to the number of arrangements with person A at seat 1 divided by 1 (since each orbit contributes one such arrangement). Wait, no. If there are M orbits, each orbit contains one arrangement with person A at seat 1, one at seat 2, one at seat 3, one at seat 4. Therefore, total number of arrangements with person A at seat 1 is M. But we know that number is 3! = 6. Hence, M = 6. Which matches the Burnside count of 6. Therefore, number of orbits is 6, and number of arrangements with person A at seat 1 is also 6. Wait, but how is this possible?

If each orbit contains one arrangement with person A at seat 1, then total number of such arrangements is equal to the number of orbits. But we also have that number of such arrangements is 3! = 6. Therefore, number of orbits is 6. Which matches. Therefore, for n=4, it works.

For n=8, similarly, the number of arrangements with person A at seat 1 is 7! = 5040. Each orbit contains exactly one such arrangement. Therefore, number of orbits is 5040. But according to Burnside's lemma, the number of orbits is 8! / 4 = 10080. Contradiction!

Wait, this suggests that my previous assumption that each orbit contains exactly one arrangement with person A at seat 1 is wrong.

Wait, but for n=4, it worked. So what's different with n=8?

Wait, let me think with n=4. If I have an orbit, it contains 4 arrangements, each with person A at different seats. Therefore, the number of orbits is equal to the number of arrangements with person A at seat 1 (which is 6) because each orbit has exactly one arrangement with person A at seat 1. Hence, number of orbits is 6.

But for n=8, if each orbit contains one arrangement with person A at seat 1, then number of orbits is 7! = 5040. But according to Burnside's lemma, it's 8! / 4 = 10080. Therefore, there's a contradiction. Therefore, my assumption that each orbit contains exactly one arrangement with person A at seat 1 must be wrong.

Wait, why? Because in n=4, each orbit has four positions for person A, but in n=8, maybe it has more?

Wait, no. For any orbit, the number of positions where person A can be is equal to the number of rotations. Wait, the orbit size is 4, so for each arrangement in the orbit, person A is in one of the 4 positions. So for each orbit, there are 4 arrangements with person A in different seats. Hence, the number of orbits should be equal to the number of arrangements with person A at seat 1 divided by 1 (since each orbit contains one such arrangement). Wait, no, if each orbit has 4 arrangements with person A in different seats, then the number of orbits is equal to the number of arrangements with person A at seat 1 multiplied by 1 (since each orbit contains one arrangement at seat 1). Wait, this is confusing.

Let me think of it as follows. Total number of arrangements is 8!. Number of orbits is 8! / 4 = 10080. Number of arrangements with person A at seat 1 is 7! = 5040. If each orbit contains 1 arrangement with person A at seat 1, then number of orbits is 5040. But 5040 ≠ 10080. Therefore, each orbit must contain 2 arrangements with person A at seat 1. But how?

Wait, this suggests that my previous assumption that each orbit has size 4 is wrong. If each orbit has size 2, then number of orbits is 8! / 2 = 20160. But this contradicts Burnside's lemma.

Alternatively, maybe some orbits have size 4 and some have smaller size. But earlier, we thought that no non-identity rotations fix any arrangement, so all orbits have size 4. So why the discrepancy?

Wait, let me try to think of a specific example. Let's take two arrangements: one where person A is at seat 1 and person B is at seat 2, and another arrangement where person A is at seat 1 and person B is at seat 3. Are these in the same orbit? Or different orbits?

If you rotate the first arrangement by 90°, person A moves to seat 3, and person B moves to seat 4. If the second arrangement has person A at seat 1 and person B at seat 3, then rotating the first arrangement 90° would give a different arrangement. So these are different orbits.

But if I have an arrangement 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 C is at seat 2, these are in different orbits.

So, if I fix person A at seat 1, each different arrangement is in its own orbit. Therefore, the number of orbits where person A is at seat 1 is equal to the number of such arrangements, which is 7! = 5040. But according to Burnside, total number of orbits is 10080, which is double. So where are the other 5040 orbits?

Ah! Wait, maybe the orbits are not just those where person A is at seat 1, but also where person A is at seat 3, 5, 7. But if we fix person A at seat 1, we are only counting orbits that contain an arrangement with person A at seat 1. But there are other orbits where person A is at seat 3, 5, or 7. However, since we're considering equivalence up to rotation, those orbits are already counted when we fix person A at seat 1. Wait, no. If we fix person A at seat 1, we are only considering arrangements where person A is at seat 1, but the other orbits where person A is not at seat 1 are not considered. Hence, the total number of orbits is equal to the number of orbits where person A is at seat 1, because by rotating, any orbit can have person A at any seat. Wait, but this would mean that the total number of orbits is equal to the number of arrangements with person A at seat 1, which is 7! = 5040. But Burnside says it's 10080.

This is very confusing. Maybe I need to reference a textbook or recall the formula for circular permutations with rotational symmetry.

Wait, in general, the number of distinct circular arrangements of n people is (n-1)! because you fix one person and arrange the rest. But in this case, the table is square, but we are considering rotations as equivalent. If the table were circular, the number would be (8-1)! = 5040. But the problem says it's a square table, so maybe the symmetry group is smaller? Wait, no, the problem says that two configurations are equivalent if one is a rotation of another. So it's still considering rotational symmetry, regardless of the table shape. So if it were a circular table, the number would be (8-1)! = 5040. But since it's a square table, with rotational symmetry of order 4, the number should be 8! / 4 = 10080. But this contradicts the circular table result.

Wait, why? Because for a circular table, the rotational symmetry group has order 8 (rotations by 360/8 = 45°), so number of distinct arrangements is 8! / 8 = 7! = 5040. For a square table, rotational symmetry group has order 4, so number of distinct arrangements is 8! / 4 = 10080. Ah! So this makes sense. The difference is the size of the symmetry group. For a circular table, we have more symmetries (rotations by 45° increments), while for a square table, only rotations by 90° increments.

Therefore, the answer depends on the symmetry group. If the problem considers rotations by any degree (i.e., circular table), the answer is 7! = 5040. If it considers rotational symmetry only to the square's symmetry (rotations by 90°), then the answer is 8! / 4 = 10080.

But the problem says "a square table with 2 people on a side". It doesn't specify whether rotations are considered equivalent only for the square's symmetry or for the full circular symmetry. However, usually, when a table is described as square, it's implied that rotations are considered equivalent only under the square's symmetries. But the problem specifically says "Two configurations are considered equivalent if one is a rotation of another."

But what is a rotation of a square table? A rotation by 90°, 180°, 270°, or 360° (identity). Therefore, the rotational symmetry group is of order 4, hence number of distinct arrangements is 8! / 4 = 10080.

But then why does fixing one person give 7! = 5040? Because fixing one person removes the rotational symmetry. Wait, but if we fix one person, we are effectively choosing a representative from each orbit. But if the orbits have size 4, then the number of orbits is 8! / 4. If I fix one person, I'm counting the number of orbits where that person is at a specific seat, which is 7!, but this would be equal to total number of orbits divided by 4 (since each orbit contains 4 different positions for that person). Hence, total number of orbits is 4 * 7! = 4 * 5040 = 20160, which contradicts Burnside's lemma result of 10080.

Wait, but this suggests that fixing one person is not the correct way to count. Hence, the correct way is Burnside's lemma, giving 10080.

But then why in the circular table case, fixing one person gives the correct result? Because in the circular table, the rotational symmetry group has order 8, so number of orbits is 8! / 8 = 7! = 5040. If you fix one person, you get 7! = 5040, which is the same as 8! / 8. Hence, it works.

Therefore, in the square table case, since the rotational symmetry group has order 4, the number of orbits is 8! / 4 = 10080. Fixing one person would not give the same result, because fixing one person removes the rotational symmetry, but the number of orbits should be equal to the number of arrangements with one person fixed. But this doesn't match.

Wait, maybe the error is in assuming that fixing one person accounts for all symmetries. But if the symmetry group has order 4, fixing one person might not fully account for it. Wait, no. Fixing one person should eliminate all rotational symmetries, as any rotation would move that person to another seat, which is fixed. Hence, fixing one person should give the correct count. But why the discrepancy?

Wait, I think the key is that in the square table case, fixing one person does not fully account for the group action. Let me think of it this way: if I fix one person at a seat, I'm effectively considering arrangements where that seat is fixed, which eliminates the rotational symmetry. Hence, the number of distinct arrangements should be equal to 7!. But according to Burnside's lemma, it's 8! / 4.

Therefore, one of these must be wrong. But in the case of the circular table, both methods agree. So why discrepancy here?

Wait, let's take a small example with n=4, but with a square table (4 seats, one per side). If we use Burnside's lemma, we get (4! + 0 + 0 + 0)/4 = 6. If we fix one person, we get 3! = 6. So it works.

Now, take n=8, but with a square table, and try to apply the same logic. If I fix one person, the number of arrangements is 7! = 5040. But according to Burnside's lemma, it's 8! / 4 = 10080.

Wait, but perhaps in this case, fixing one person is not the correct way because the square table has more symmetries? No, it's only rotations.

Wait, let me think about what fixing one person means. If I fix person A at seat 1, then rotations that move seat 1 are no longer allowed. Hence, the number of arrangements should be 7! = 5040. But according to Burnside's lemma, it's 10080. There must be a miscalculation in the Burnside's lemma part.

Wait, let me go back to the Burnside's lemma. For the square table with 8 seats, group has 4 rotations: 0°, 90°, 180°, 270°. Fixed arrangements:

- 0°: 8! = 40320

- 90°: 0

- 180°: ?

Wait, earlier I thought that fixed arrangements under 180° is 0, but maybe it's not?

Wait, if an arrangement is fixed under 180° rotation, then each person is seated opposite to themselves. But since the table has 8 seats, the opposite seat of seat i is seat i + 4. So for the arrangement to be fixed under 180° rotation, the person at seat i must be the same as person at seat i + 4. But since all people are distinct, this is only possible if each pair of opposite seats has the same person, which is impossible. Hence, fixed arrangements under 180° is 0.

But wait, what if the arrangement is such that person A is at seat 1 and seat 5, but since people are distinct, this is impossible. Hence, fixed arrangements under 180° is 0.

So Burnside's lemma gives (40320 + 0 + 0 + 0)/4 = 10080.

But how can this be? If fixing one person gives 5040, which is half of 10080? Is there a possibility that the problem considers rotations and reflections as equivalent? But the problem says only rotations.

Wait, if we considered reflections as well, the group would have 8 elements (rotations and reflections), and the count would be different. But the problem only mentions rotations.

Wait, maybe the error is that when we fix one person, we are fixing a specific seat, but on a square table, there are symmetries that could still be present. Wait, no. If you fix a person at a specific seat, you eliminate all rotational symmetries.

Wait, let's think about it in terms of the orbit. If we have an arrangement with person A fixed at seat 1, then its orbit includes all rotations of this arrangement. But if person A is fixed at seat 1, then none of the rotations can move person A, so the orbit contains only the arrangement itself. Hence, fixing one person should give the correct count, as each orbit is represented once. Hence, the number of orbits is 7! = 5040.

But this contradicts Burnside's lemma result. Where is the error?

Ah! I think the error is in the application of Burnside's lemma. The group actions are on the seats, but when we fix a seat, we are not considering the group action. Wait, let me recall: Burnside's lemma counts the number of orbits under the group action. In this case, the group is rotations of the table, which act on the set of seatings.

If we fix one person at a seat, we are effectively choosing a representative for each orbit by fixing that person at that seat. Hence, the number of such representatives is equal to the number of orbits.

But if fixing one person gives 7! = 5040, then the number of orbits is 5040, and Burnside's lemma must have been applied incorrectly.

But why did Burnside's lemma give 10080? Because I thought fixed arrangements under rotations were 0, but maybe there are fixed arrangements under 180°?

Wait, let me think again about fixed arrangements under 180° rotation. Suppose we have an arrangement fixed under 180° rotation. This means that the person at seat 1 is the same as the person at seat 5, seat 2 same as seat 6, etc. But since all people are distinct, this is impossible. Therefore, fixed arrangements under 180° is 0.

Wait, but what if the table has an even number of seats and we have pairs? For example, if the table has 4 seats, and we have an arrangement fixed under 180° rotation, it requires pairs of people. But since all people are distinct, this is impossible. Hence, fixed under 180° is 0.

So why is there a discrepancy?

I think the key is that when we fix one person, we are not accounting for the fact that some rotations might have fixed points. But wait, no, we already considered that.

Wait, let me check with n=4 again. Using Burnside's lemma, we get 6. Fixing one person gives 6. So it works. For n=8, why not?

Wait, let's try to calculate the number of orbits using the formula for circular tables. If the table were circular, with 8 seats, the number of arrangements would be 7! = 5040. But if it's a square table, why is it different? Because for a circular table, the group has 8 elements (rotations by 45°, 90°, ..., 315°), while for a square table, the group has 4 elements (rotations by 90°, 180°, 270°, 360°). Therefore, for the square table, the number of distinct arrangements is 8! / 4 = 10080, while for the circular table, it's 7! = 5040.

Therefore, the answer depends on the symmetry group. Since the problem says it's a square table, we should use the square's symmetry group. Hence, the answer is 8! / 4 = 10080.

But why does fixing one person give a different answer? Because fixing one person assumes that the symmetry group is of size 8, but in reality, the symmetry group is of size 4. Hence, fixing one person is not the correct method here.

Wait, but how? If I fix one person, I am removing the rotations. If the symmetry group has size 4, then fixing one person should give 8! / 4 = 10080. But no, fixing one person is different.

Wait, let me think. If the symmetry group has size 4, and we fix one person, how many arrangements are there? It should be total number of arrangements divided by the size of the group. Wait, no.

Actually, the number of orbits is equal to the total number of arrangements divided by the size of the group, if all orbits have the same size. Which is the case here, since no non-identity element fixes any arrangement. Hence, number of orbits is 8! / 4 = 10080.

But if I fix one person, I should get 10080, but I get 7! = 5040. This suggests that fixing one person is not the correct way here.

Wait, but in the circular table example, fixing one person works. Why does it work there? Because in the circular table, fixing one person is equivalent to dividing by the size of the group (which is 8), giving 8! / 8 = 7!.

For the square table, if we fix one person, it's equivalent to dividing by the size of the group (4)? But why does this give 8! / 4 = 10080, but fixing one person gives 7! = 5040? It doesn't.

Wait, no. If you fix one person, you are not dividing by the group size, but fixing a position. How is this related?

Let me think of it as follows: the number of distinct arrangements is equal to the number of ways to arrange the people divided by the number of rotational symmetries, if each orbit has size equal to the number of symmetries. Hence, it's 8! / 4 = 10080.

On the other hand, fixing one person gives 7! = 5040, which is 8! / 8. This suggests that fixing one person assumes a larger symmetry group (8), but in reality, the symmetry group is 4. Hence, fixing one person is not the correct approach here.

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

But why does fixing one person work for the circular table? Because in the circular table, the symmetry group is size 8, and fixing one person is equivalent to dividing by 8, giving 8! / 8 = 7! = 5040, which matches.

For the square table, if we want to use the same logic, we should divide by 4, giving 8! / 4 = 10080.

Therefore, the correct answer is 10080, and the confusion with fixing one person is because fixing one person assumes a different symmetry group.

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

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

To solve the problem of seating 8 people around a square table with **2 people on each side**, and considering that **rotational symmetry makes some configurations equivalent**, we proceed as follows:

---

### **Understanding the Setup**

- The table has **4 sides**, and **2 seats per side**, so **8 total seats** are available.
- We are to place **8 distinct individuals** in these 8 seats.
- **Rotations** that result in the same seating layout are considered **equivalent**: if you rotate the table by 90°, 180°, or 270°, those are not counted as new configurations.

Thus, we are counting **distinct seating arrangements up to rotation**.

---

### **Identifying the Symmetry Group**

The key is to understand which **rotational symmetries** are relevant. A square has **rotational symmetry of order 4**, meaning that rotating the table by 90°, 180°, 270°, or 360° (identity) gives equivalent configurations. So, the group of symmetries (rotations) has **4 elements**.

---

### **Using Burnside's Lemma to Count Distinct Arrangements**

Burnside's Lemma is ideal for this type of problem: it tells us that the number of distinct arrangements (under group action) is the **average number of arrangements fixed** by each group element.

Let’s define the group:
- Identity (0° rotation): Fixes all arrangements.
- Rotation by 90°: No arrangement is fixed (because a 90° rotation maps each seat to a new one, requiring all seats to have the same person, which is impossible since all people are distinct).
- Rotation by 180°: Similarly, no arrangement is fixed (each person is opposite someone else; again, conflict with distinctness).
- Rotation by 270°: Same reasoning as 90°.

So, the only group element that fixes any arrangement is the identity. Therefore:

- Number of arrangements fixed by identity: $ 8! = 40320 $
- Number of arrangements fixed by 90°, 180°, and 270°: 0 each

Now, applying Burnside’s Lemma:

$$
\text{Number of distinct arrangements} = \frac{1 \times 8! + 0 + 0 + 0}{4} = \frac{40320}{4} = \boxed{10080}
$$

---

### **Why Fixing a Person Isn't Directly Applicable Here**

A common technique in circular permutation is to *fix one person* to account for rotation. But this works *only* if the group of symmetries is of size *equal to the number of seats*, which is the standard circular symmetry (rotating 360° / n).

In our case, the group size is **4** (due to the square's symmetry), not 8. Hence, fixing one person and dividing by that number (or directly calculating $7!$) is **not** appropriate here. Doing so introduces a conceptual mismatch with the actual group of symmetries.

Thus, the **correct** and **consistent** method to apply is **Burnside’s Lemma**, which considers the symmetry group of the square table accurately.

---

### **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 configuration. Hmm, let me think.

First, I remember that arranging people around a table usually involves circular permutations. But this is a square table, so maybe there's a difference? For a circular table, the number of arrangements is (n-1)! because rotations are considered the same. But a square table, even though it's a polygon, might have different symmetries? Wait, but the problem says that rotations are considered equivalent. So maybe it's similar to a circular table but with some restrictions?

Wait, a square table has four sides. If there are two people on each side, that means each side has two seats? So total seats would be 4 sides * 2 people = 8 people. Yep, that matches the question. So, it's like a square with seats on each side, two per side.

But how does that affect the number of arrangements? For circular tables, we fix one person's position to account for rotations. But here, since it's a square, maybe rotations by 90 degrees would be considered equivalent? Wait, the problem says "rotations of another", so maybe rotations by any multiple of 90 degrees? Let me check.

If we rotate the table 90 degrees, the positions of the people would shift. But the problem states that two configurations are equivalent if one is a rotation of another. So, for example, if you rotate the table by 90 degrees, the arrangement is considered the same. Therefore, similar to circular permutations, but the number of rotational symmetries is different. For a square, there are four rotational symmetries: 0°, 90°, 180°, and 270°. So, the number of rotational symmetries is 4. Therefore, maybe the formula is similar to circular permutations but divided by 4 instead of n?

Wait, for a circular table with n seats, the number of distinct arrangements is (n-1)! because there are n rotations that result in equivalent configurations. But here, since there are only 4 rotational symmetries, maybe the formula is (number of linear arrangements)/4? Let me think.

Alternatively, maybe I should fix one person's position to account for rotations. If I fix one person at a specific seat, then arrange the others. For example, fix one person at a specific seat, then arrange the remaining 7 people. But wait, if the table is square, fixing one person might fix the rotational symmetry?

Wait, but if we fix one person at a particular seat, then rotations are already accounted for. For example, in a circular table, fixing one person removes the rotational symmetry. Similarly, for the square table, if I fix one person at a seat, then the rest can be arranged relative to that person. However, the square also has reflections, but the problem doesn't mention reflections being equivalent. It only mentions rotations. So reflections would be considered different.

So maybe the formula is similar to circular permutations but divided by 4? Let me check with a smaller number. Suppose there are 4 people, one on each side of a square table. How many arrangements? If we consider rotations equivalent, then it's (4-1)! = 6? Wait, but if there are four seats, and rotations are considered equivalent, the number is (4-1)! = 6. But if we fix one person, then we have 3! = 6. So that works. However, if we consider the square's rotational symmetry, there are 4 rotations. So in general, for n seats around a table with rotational symmetry of order k, the formula is n!/k? Wait, for circular table, n seats, rotational symmetry of order n, so number of arrangements is n!/n = (n-1)!.

But here, with the square table, if there are 8 seats, and rotational symmetry of order 4, then the number of arrangements would be 8!/4? But is that correct?

Wait, let me think again. Suppose we have a square table with 8 seats: two on each side. Let me imagine the seats as follows: seat 1 and seat 2 on the north side, seat 3 and seat 4 on the east side, seat 5 and seat 6 on the south side, seat 7 and seat 8 on the west side. Then, rotating the table 90 degrees clockwise would move seat 1 to seat 3's position, seat 2 to seat 4, seat 3 to seat 5, seat 4 to seat 6, seat 5 to seat 7, seat 6 to seat 8, seat 7 to seat 1, seat 8 to seat 2. So, each rotation moves each seat to the next side's position.

So if we consider two seatings equivalent if one can be rotated to get the other, then how many distinct arrangements are there?

In general, the formula for counting distinct arrangements under rotational symmetry is given by Burnside's lemma, which averages the number of fixed points of the group actions. But maybe that's overcomplicating? Wait, but if I can just divide by the number of symmetries?

Wait, but Burnside's lemma is more reliable here. Let me recall. Burnside's lemma says that the number of distinct arrangements is equal to the average number of elements fixed by each group action. So, for the group of rotations, which has 4 elements (0°, 90°, 180°, 270°), we need to compute the number of arrangements fixed by each rotation, then add them up and divide by 4.

Alternatively, if all rotations except the identity have no fixed arrangements, then the total number would be (total arrangements + 0 + 0 + 0)/4 = total arrangements /4. But is that the case?

Wait, let me check. For the identity rotation (0°), all arrangements are fixed, so that's 8!.

For the 90° rotation, how many arrangements are fixed? That would mean that rotating the table 90° leaves the arrangement unchanged. So, each person must be seated in a position that maps to themselves when rotated. But since the rotation cycles all 8 seats in a single cycle? Wait, no. Let me think.

If you rotate the table 90°, then each seat moves to the next seat on the adjacent side. For example, seat 1 goes to seat 3, seat 3 goes to seat 5, seat 5 goes to seat 7, seat 7 goes to seat 1. Similarly, seat 2 goes to seat 4, seat 4 goes to seat 6, seat 6 goes to seat 8, seat 8 goes to seat 2. Therefore, the rotation decomposes the seats into two cycles of length 4: one cycle for the odd-numbered seats (1,3,5,7) and one cycle for the even-numbered seats (2,4,6,8). For an arrangement to be fixed under a 90° rotation, the people in each cycle must be the same. But since all people are distinct, this is only possible if all people in each cycle are the same. But since all people are distinct, this is impossible. Therefore, there are 0 fixed arrangements under 90° rotation.

Similarly, for the 270° rotation, same logic applies. It's equivalent to rotating -90°, which also cycles the seats in the same manner, so same result: 0 fixed arrangements.

Now for the 180° rotation. This rotation swaps seat 1 with seat 5, seat 2 with seat 6, seat 3 with seat 7, seat 4 with seat 8. So, each person must be seated opposite their original position. Therefore, for an arrangement to be fixed under 180° rotation, each person must be seated opposite to themselves, meaning that each pair (1 and 5, 2 and 6, etc.) must have the same person. But again, since all people are distinct, this is only possible if each pair has the same person, which is impossible. Wait, hold on, no. Wait, if you have person A at seat 1, then person A must also be at seat 5. But since all people are distinct, you can't have two people at the same seat. Therefore, actually, there are no fixed arrangements under 180° rotation as well? Wait, that can't be right. Wait, maybe I made a mistake here.

Wait, no. Wait, the arrangement is fixed under 180° rotation if rotating the table 180° results in the same arrangement. That would mean that person at seat 1 is the same as person at seat 5, etc. But since seats are occupied by different people, this would require that person at seat 1 is the same as person at seat 5, which can't happen unless they are the same person. But since all people are distinct, there are no such arrangements. Therefore, fixed arrangements under 180° rotation would be zero? Wait, but that would mean that Burnside's lemma gives us (8! + 0 + 0 + 0)/4 = 8!/4. But is this correct?

But that seems conflicting with my intuition. Wait, for example, with n=4 seats at a square table, with one seat per side. Let's do that case. If there are 4 seats, one on each side. How many arrangements considering rotations equivalent?

By Burnside's lemma: group has 4 rotations. Identity fixes 4! = 24. Rotations by 90° and 270° fix 0 arrangements (since each cycle is length 4, requires all seats to have same person). Rotation by 180°: fixes arrangements where opposite seats are same. So how many? For 4 seats, opposite seats must be same. But since people are distinct, we need to pair them. Wait, but if you have 4 people, and you need to assign them to seats such that seat 1 = seat 3 and seat 2 = seat 4. So the number of such arrangements is 4! / (2! * 2!) ? Wait, no. Wait, actually, you have to assign people to pairs. So, first choose a person for seat 1 and 3, then a person for seat 2 and 4. So, it's 4 choices for seat 1, then 3 choices for seat 2, and then the rest are determined. But since seat 1 and 3 are the same, but they are different seats. Wait, actually, the number of arrangements fixed under 180° rotation would be the number of ways to assign people such that seat 1 and seat 3 are same, seat 2 and seat 4 are same. But since people are distinct, you can't have two people in the same seat. Therefore, actually, there are zero such arrangements? Wait, that can't be. Wait, maybe I need to think differently.

Wait, if the arrangement is fixed under 180° rotation, then person in seat 1 must be the same as person in seat 3. But since each seat is occupied by a different person, this is impossible. Therefore, there are zero arrangements fixed under 180° rotation. Therefore, Burnside's lemma gives (24 + 0 + 0 + 0)/4 = 6. Which is equal to (4-1)! = 6. So that works. Therefore, for n=4, this formula works. So maybe in the original problem with 8 seats, the Burnside's lemma gives 8! / 4. Which would be 40320 / 4 = 10080. But that seems high?

Wait, but let's check with another example. Suppose n=2 seats on a square table? Wait, that's not possible, since square has four sides. Maybe n=2 people? Hmm, not sure. Maybe n=8 is okay.

But wait, if we fix one person, then arrange the rest, how does that work?

If we fix one person at a seat, then the remaining 7 can be arranged in 7! ways. But since rotations are considered equivalent, fixing one person accounts for rotational symmetry. However, in the square case, fixing one person would eliminate rotational symmetry? Wait, for example, suppose I fix person A at seat 1. Then, rotations that move seat 1 to another seat are eliminated. Therefore, the number of distinct arrangements would be 7!.

But earlier Burnside's lemma suggested 8! / 4 = 10080, which is 8!/4, and 7! is 5040. Which is different. So which is correct?

Wait, there's a contradiction here. So which one is correct?

Wait, let me think again. For circular table with n seats, fixing one person gives (n-1)! arrangements. For the square table with 8 seats, if we fix one person, we are effectively removing rotational symmetry? But the square has rotational symmetry of order 4, so fixing one person might not fully account for all symmetries? Wait, no. If you fix one person at a specific seat, then rotations that move that seat are eliminated. So for example, if you fix person A at seat 1, then the other rotations would move seat 1 to other positions, but since we've fixed person A there, those rotations are not allowed. Therefore, the number of distinct arrangements is 7!.

But according to Burnside's lemma above, it was 8!/4 = 10080, which is 8!/4 = 40320 / 4 = 10080, whereas 7! is 5040. These are different. So which is correct?

Wait, there must be an error in one of the approaches. Let me think again about Burnside's lemma. For the square table with 8 seats, the group of rotations is of order 4: rotations by 0°, 90°, 180°, 270°. Let me recalculate the fixed arrangements.

For identity rotation: all 8! arrangements are fixed.

For rotation by 90°: an arrangement is fixed if rotating 90° leaves it unchanged. As discussed earlier, this would require that each person is mapped to themselves under rotation. However, the rotation decomposes the seats into two cycles of length 4: one cycle for the seats on the "odd" positions (1,3,5,7) and another cycle for the "even" positions (2,4,6,8). For the arrangement to be fixed, the person in seat 1 must be the same as seat 3, seat 3 same as seat 5, etc. Since all people are distinct, this requires that all these seats have the same person, which is impossible. Therefore, fixed arrangements under 90° rotation: 0.

Similarly for 270° rotation: same result, 0.

For rotation by 180°: each seat is mapped to its opposite seat. So seat 1 swaps with 5, 2 swaps with 6, 3 swaps with 7, 4 swaps with 8. Therefore, for the arrangement to be fixed, the person in seat 1 must be the same as seat 5, person in seat 2 same as seat 6, etc. Again, since all people are distinct, this is impossible. Therefore, fixed arrangements under 180° rotation: 0.

Therefore, total fixed arrangements: 8! + 0 + 0 + 0 = 40320. Divided by 4, gives 40320 / 4 = 10080. But if I fix one person, I get 7! = 5040. Which one is correct?

Wait, if we fix one person, we need to consider that fixing a person removes the rotational symmetry. However, in the case of the square table, maybe fixing one person is not sufficient? Wait, no. If you fix person A at seat 1, then rotations that move seat 1 to other positions are no longer allowed. Therefore, the number of distinct arrangements should be 7!.

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

Wait, let's take a small example again. Suppose n=4 seats, one per side. Then, according to Burnside's lemma, it would be (4! + 0 + 0 + 0)/4 = 24 / 4 = 6. Which is (4-1)! = 6. So that works. Now, if we fix one person, say seat 1, then we have 3! = 6, which is same as Burnside's result. So in that case, both methods agree.

Wait, so why in the case of n=8 seats, fixing one person gives 7! but Burnside's lemma gives 8! / 4? Because 7! = 5040 and 8! / 4 = 10080. They are different. So that suggests that there is an error in one of the reasonings.

Wait, let me think again about Burnside's lemma for n=8 seats. If I fix one person, then I can think of it as dividing by the number of rotations. But in the case where the number of rotations is 4, then dividing by 4 would be the same as fixing one position?

Wait, but in circular table, we have n rotations. So to fix one position, divide by n. Here, for square table, number of rotations is 4, so divide by 4. But in the case of n=4 seats, dividing by 4 gives (4-1)! = 6. Which is same as fixing one person. So maybe here, with 8 seats, fixing one person is equivalent to dividing by 4? But how?

Wait, if I fix one person, then the number of arrangements is 7! = 5040. According to Burnside's lemma, it's 8!/4 = 40320 / 4 = 10080. Which is double. Therefore, something is wrong here.

Wait, perhaps I made a mistake in Burnside's lemma calculation. Let me check again.

Wait, when we fix a person, we are effectively considering arrangements up to rotation. But perhaps in the case of the square table, fixing one person doesn't fully account for the symmetries? Wait, no. Fixing one person should eliminate all rotational symmetries.

Alternatively, maybe the Burnside's lemma was applied incorrectly. Let me re-examine.

Burnside's lemma says the number of distinct arrangements is equal to (number of arrangements fixed by each group element averaged over the group). So group has 4 elements: identity, rotation 90°, 180°, rotation 270°.

Number fixed by identity: 8!.

Number fixed by 90°: as before, 0.

Number fixed by 180°: let's re-examine.

If we have an arrangement fixed under 180° rotation, that means that each person is seated opposite to themselves. Wait, no. It means that when you rotate the table 180°, the arrangement looks the same. Therefore, the person at seat 1 must be the same as the person at seat 5. But since all people are distinct, this can't happen. Therefore, number of fixed arrangements under 180° rotation is 0? But that can't be, because maybe there are arrangements where people are arranged in pairs?

Wait, no. If the person at seat 1 is person A, then after rotation, person A must be at seat 5. But seat 5 is a different seat. Therefore, unless person A is sitting at both seat 1 and seat 5, which is impossible, there can be no fixed arrangements. Hence, fixed arrangements under 180° rotation is 0. Then Burnside's lemma gives (8! + 0 + 0 + 0)/4 = 8!/4. However, if we fix one person, we get 7!.

This discrepancy suggests that one of the methods is wrong. Wait, maybe fixing a person doesn't account for the entire symmetry?

Wait, take n=4. With Burnside's lemma, we get (4! + 0 + 0 + 0)/4 = 6. Which is same as fixing one person (3! = 6). So in that case, it works. What about with n=8 seats?

Wait, suppose we fix one person. Let me think of the square table with 8 seats. If we fix person A at seat 1, then the remaining 7 seats can be filled in 7! ways. However, are there still rotational symmetries left? For example, rotating the table 180°, which would swap seat 1 with seat 5. But seat 1 is fixed with person A, so seat 5 must also have person A. But that would require person A to be in two seats, which is impossible. Therefore, fixing person A eliminates all rotational symmetries? Because any rotation would require some person to be in multiple seats, which is impossible. Therefore, fixing one person should give the correct number of arrangements, which is 7!.

But according to Burnside's lemma, it's 8! / 4. Which is different. So there must be an error in Burnside's calculation. Wait, let's check Burnside's lemma again for n=8 seats.

Wait, perhaps my analysis of fixed arrangements under 180° is wrong. Let me think again. If we have an arrangement fixed under 180° rotation, that means that person at seat i is the same as person at seat i + 4 (modulo 8). However, since all seats are distinct, this would require that each pair of opposite seats has the same person. But since people are distinct, this is only possible if each pair is assigned the same person, which is impossible unless you have duplicate people. Therefore, there are zero such arrangements. Hence, Burnside's lemma calculation is correct?

But why is there a discrepancy?

Wait, maybe the group action isn't just rotations? Wait, the problem says "rotations of another", so only rotations. So group is cyclic group of order 4. But perhaps my mistake is in considering the seats as labeled? Wait, if the seats are unlabeled, then Burnside's lemma applies. But if seats are labeled, then the count is different. Wait, in our case, are the seats labeled? Or are they considered unlabeled?

Wait, when we talk about seating arrangements, the seats are considered distinct because they are different positions around the table. However, when considering rotations, we consider two arrangements equivalent if one can be rotated to get the other. So in Burnside's lemma, we need to count colorings (arrangements) up to rotation. So if we have 8 seats, each seat can be assigned a distinct person. So it's like coloring the seats with 8 different colors, considering rotations equivalent.

So Burnside's lemma says the number is (fixed by identity + fixed by 90 + fixed by 180 + fixed by 270)/4.

As above, fixed by identity is 8!.

Fixed by 90 and 270: 0.

Fixed by 180: ?

Wait, maybe my previous conclusion that fixed by 180 is zero is wrong. Wait, suppose we have an arrangement fixed under 180° rotation. That means that the person at seat 1 is same as seat 5, seat 2 same as seat 6, etc. However, since each person is unique, this would require that person at seat 1 equals person at seat 5, which is impossible. Therefore, fixed arrangements under 180° is zero. Therefore, Burnside's lemma gives 8! / 4.

But then why does fixing one person give a different answer? There must be something wrong here.

Wait, let's think of fixing person A at seat 1. Then, the rest can be arranged freely. However, in this case, are we accounting for all rotations? Suppose we don't fix person A, but consider rotations. So the total number of arrangements is 8!. Each equivalence class under rotation has size equal to...?

In general, for a group action, the number of equivalence classes is equal to the total number of arrangements divided by the size of the group, if all orbits have the same size. However, this is only true if no non-identity element fixes any arrangement. Which is the case here, because we saw that only identity fixes arrangements. Therefore, the number of equivalence classes is 8! / 4. Hence, Burnside's lemma gives 8! / 4. Therefore, fixing one person must not be the correct approach here?

Wait, but how? If I fix one person, then I should be able to count the number of equivalence classes. Wait, fixing one person is equivalent to considering orbits where one seat is fixed. Wait, maybe not. Let me think.

If we fix one person, then we are essentially choosing a representative from each orbit by fixing that person in that seat. But the number of orbits would be equal to the number of such arrangements. However, if fixing one person gives 7! arrangements, then the total number of orbits should be 7! = 5040, which is different from Burnside's count of 8! / 4 = 10080. Therefore, clearly, one of the logic is wrong.

Wait, here's a key point. When you fix one person, you are effectively counting the number of distinct arrangements where that person is in a specific seat. However, in the Burnside's lemma, we count all equivalence classes, where each equivalence class can have different numbers of arrangements. Wait, but if all orbits have size equal to the group size, which is 4, then the number of orbits is 8! / 4. However, if there are some orbits with smaller size, then Burnside's lemma counts them properly.

But earlier, when we considered fixed arrangements under non-identity rotations, we found that there are none. Therefore, all orbits have size equal to the group size, which is 4. Therefore, number of orbits is 8! / 4. Therefore, fixing one person should not give you the same number. Wait, but how?

Wait, the number of orbits is equal to the number of distinct arrangements under rotation. If I fix one person, then how does that relate?

Suppose I fix person A at seat 1. Then, the number of arrangements where person A is at seat 1 is 7!. However, each orbit contains exactly 4 different arrangements (rotated versions). However, some of these orbits may contain arrangements where person A is at seat 1, seat 3, seat 5, or seat 7. Wait, no. If an orbit contains an arrangement where person A is at seat 1, then rotating it by 90°, 180°, 270°, would move person A to seat 3, seat 5, seat 7. Therefore, each orbit that contains an arrangement with person A at seat 1 also contains arrangements with person A at seats 3, 5, 7. Therefore, the number of orbits where person A is at seat 1 is equal to the total number of orbits divided by 4 (since each orbit has 4 positions where person A can be). Therefore, total number of orbits is 4 * (number of arrangements with person A at seat 1) = 4 * 7! = 4 * 5040 = 20160. But according to Burnside's lemma, total number of orbits is 8! / 4 = 40320 / 4 = 10080. Which is half of 20160. So that suggests inconsistency.

Therefore, there is a contradiction here. Where is the mistake?

Wait, maybe the orbits have size 4, but the number of orbits is 8! / 4 = 10080. Therefore, the number of orbits where person A is at seat 1 is 10080 / 4 = 2520. But according to fixing person A at seat 1, we have 7! = 5040. So 2520 vs 5040. So there is a factor of 2 difference. Therefore, something is wrong here.

Wait, perhaps orbits are not size 4. Maybe some orbits have smaller size? But earlier we said that no non-identity element fixes any arrangement, so by orbit-stabilizer theorem, the size of each orbit is equal to the size of the group divided by the size of the stabilizer subgroup. Since the stabilizer is trivial (only identity), the size of each orbit is 4. Therefore, if there are 10080 orbits, each of size 4, the total number of arrangements is 10080 * 4 = 40320 = 8!, which is correct. Therefore, the number of orbits where person A is at seat 1 is equal to total number of orbits multiplied by the probability that a random arrangement has person A at seat 1. Since there are 8 seats, the probability is 1/8. Therefore, the number of such orbits is 10080 * (1/8) = 1260. But according to fixing person A at seat 1, we have 7! = 5040. Which is not equal to 1260. Therefore, clearly something is wrong here.

Wait, perhaps I have misunderstood the orbit-stabilizer theorem? Let me recall. Orbit-stabilizer theorem says that the size of the orbit of an element is equal to the size of the group divided by the size of the stabilizer. In our case, the stabilizer of an arrangement is the set of rotations that leave it unchanged. Since we have no non-identity rotations that stabilize any arrangement, the stabilizer is trivial, so the orbit has size 4. Therefore, the number of orbits is 8! / 4 = 10080. However, if we fix person A at seat 1, we are counting the number of arrangements with person A at seat 1, which is 7!.

But how does that relate to the number of orbits?

Each orbit contains 4 arrangements, each with person A at different seats. If we consider all orbits, each orbit contains exactly one arrangement with person A at seat 1, seat 3, seat 5, seat 7. Wait, but if person A is at seat 1, then after rotation, they can be at seat 3, 5, 7. So for each orbit that contains an arrangement with person A at seat 1, there are 4 different arrangements in the orbit with person A at different seats. Therefore, the number of such orbits is equal to the number of arrangements with person A at seat 1 divided by 4. Wait, but if we have 7! arrangements with person A at seat 1, then the number of orbits containing such arrangements is 7! / 4? But how?

Wait, no. Each orbit has exactly 4 arrangements. If an orbit contains an arrangement with person A at seat 1, then it contains 4 such arrangements, each with person A at different seats. Therefore, the number of orbits that contain an arrangement with person A at seat 1 is equal to the number of arrangements with person A at seat 1 divided by 4, but since each orbit is counted once for each seat that person A can be in. Wait, this is getting too convoluted.

Wait, let's think of it this way. Total number of arrangements with person A at seat 1 is 7!. Each such arrangement is in an orbit of size 4. However, in each orbit, there are 4 arrangements, each with person A at seat 1, seat 3, seat 5, seat 7. Therefore, the number of orbits that contain an arrangement with person A at seat 1 is equal to the total number of arrangements with person A at seat 1 divided by 4. Wait, no. Each orbit has exactly one arrangement with person A at seat 1, one at seat 3, etc. So the total number of orbits is equal to the number of arrangements with person A at seat 1 multiplied by 4? No, that would overcount.

Wait, I think I need to use the orbit-stabilizer theorem again. The total number of arrangements with person A at seat 1 is 7!. Each orbit that contains such an arrangement has size 4. However, each orbit contains exactly one arrangement where person A is at seat 1, one at seat 3, one at seat 5, one at seat 7. Therefore, the number of orbits that contain an arrangement with person A at seat 1 is equal to the number of such arrangements divided by 4? No, it's equal to the number of such arrangements divided by the number of arrangements per orbit. Wait, each orbit contains exactly one arrangement with person A at seat 1. Wait, no. If you have an orbit, and person A is in one of the seats, then rotating the arrangement would move person A to different seats. So if an orbit contains an arrangement with person A at seat 1, it must also contain arrangements with person A at seat 3, seat 5, seat 7. Therefore, each orbit that contains person A at any seat contains 4 arrangements with person A at different seats. Therefore, the number of orbits that contain person A at seat 1 is equal to the number of orbits that contain person A in any seat, divided by 4. But the total number of arrangements with person A at any seat is 8 * 7! / 8 = 7! (wait, no). Wait, the total number of arrangements where person A is at any seat is 8 * 7! / 8 = 7!? No, total number of arrangements is 8!. The number of arrangements where person A is at seat 1 is 7!; same for any seat. Therefore, total number of arrangements where person A is at any seat is 8 * 7! / 8 = 7!? No, that doesn't make sense. Wait, total number of arrangements where person A is at some specific seat is 7!. Since there are 8 seats, the total number of arrangements where person A is at any seat is 8 * 7! / 8? No, that's not. Actually, the total number is 8 * 7! but this counts each arrangement multiple times. Wait, no. Each arrangement has person A at exactly one seat. Therefore, total number of arrangements where person A is at any seat is just equal to the total number of arrangements, which is 8!. But this counts each arrangement once. So, the number of arrangements where person A is at seat 1 is 7!, and same for other seats.

Therefore, the total number of arrangements where person A is at seat 1 is 7!. Each such arrangement is in an orbit of size 4. How many orbits are there? Each orbit contains 4 arrangements. But how many of them have person A at seat 1?

If there are N orbits, then the number of arrangements where person A is at seat 1 is equal to N multiplied by the number of arrangements in each orbit that have person A at seat 1. Since each orbit contains exactly one arrangement with person A at seat 1 (wait, no). If you have an orbit, and person A is in one of the seats, then rotating it would place person A in 4 different seats. So each orbit contains exactly 4 arrangements where person A is in different seats. Therefore, if we count all arrangements where person A is at seat 1, it is equal to N multiplied by (number of orbits containing such an arrangement) * 1 (since each orbit contributes one such arrangement). Wait, this is confusing.

Alternatively, if total number of arrangements with person A at seat 1 is 7!, and each orbit contains exactly 4 such arrangements (one for each position), then total number of orbits is 7! / 4. But total number of orbits is also equal to 8! / 4 according to Burnside's lemma. Therefore:

7! / 4 = 8! / 4 → 7! = 8! → 7! = 8 * 7! → 1 = 8, which is false. Therefore, contradiction. Hence, my assumption is wrong.

Therefore, my previous assumption that each orbit contains exactly 4 arrangements with person A in different seats is wrong.

Wait, maybe not. Suppose I have an orbit containing an arrangement where person A is at seat 1. Then, rotating it by 90°, person A moves to seat 3, by 180° to seat 5, by 270° to seat 7. Hence, in this orbit, there are four arrangements: person A at seat 1, 3, 5, 7. Therefore, the number of orbits that contain an arrangement with person A at seat 1 is equal to the number of such arrangements divided by 4. But the number of such arrangements is 7! (arrangements with person A at seat 1). Therefore, the number of orbits is 7! / 4. But according to Burnside's lemma, it should be 8! / 4 = 2 * 7!.

So, where is the problem? It seems that there are two different results. Therefore, one of the reasonings must be wrong.

Wait, the problem is that when I fix person A at seat 1, I am considering arrangements where person A is fixed, and then others are permuted. However, in Burnside's lemma, we count all arrangements up to rotation. So, if I fix person A at seat 1, I am effectively choosing a representative from each orbit where person A is at seat 1. However, the total number of orbits is equal to the number of such representatives times the number of seats person A can be in per orbit. But if each orbit contains person A in four different seats, then number of orbits is (number of representatives) * 4. Wait, no. If I have M orbits, each containing 4 arrangements where person A is in different seats, then the number of arrangements with person A at seat 1 is M * 1 (since each orbit contributes one arrangement where person A is at seat 1). Therefore, M = number of arrangements with person A at seat 1 = 7! Therefore, number of orbits is 7! But this contradicts Burnside's lemma result of 8! / 4. So I need to resolve this.

Let me use concrete numbers with n=4. For n=4 seats (one per side), Burnside's lemma says 4! / 4 = 6. If I fix person A at seat 1, then number of arrangements is 3! = 6. Which matches. But according to the previous logic, if I fix person A at seat 1, number of orbits is 3! = 6, which matches Burnside's result. So why in the case of n=8, it doesn't match?

Wait, let me try n=2. Wait, n=2? Not sure. Maybe n=8 is a different case.

Wait, let me think again. For n=4, fixing person A at seat 1 gives number of arrangements 3! = 6. Burnside gives 4! / 4 = 6. Same result. For n=8, if we fix person A at seat 1, we get 7! = 5040. If Burnside's lemma gives 8! / 4 = 10080. Why the discrepancy?

Wait, perhaps when n is divisible by the number of rotations? In the case of n=4, the number of seats is equal to the number of rotations? No, n=4 seats, rotations of order 4. For n=8 seats, rotations of order 4. So why the difference?

Wait, let's try applying Burnside's lemma for n=4, but with the same logic. For n=4 seats, rotations by 90°, 180°, 270°, identity.

Fixed by identity: 4!

Fixed by 90° and 270°: 0.

Fixed by 180°: number of arrangements fixed under 180° rotation. For n=4, this would be arrangements where person at seat 1 equals seat 3 and seat 2 equals seat 4. Since people are distinct, this is 0? But wait, no! Wait, for n=4, if you fix under 180° rotation, you need the arrangement to be symmetric under 180°. So person at seat 1 must be same as seat 3, and person at seat 2 same as seat 4. But since all people are distinct, this is impossible. Hence, fixed arrangements under 180° is 0. Then Burnside gives (24 + 0 + 0 + 0)/4 = 6, which equals (4-1)! = 6. But if we fix one person, we get 3! = 6. So it works.

But for n=8, Burnside says 8! / 4, but fixing person gives 7! = 5040. Let's compute 8! / 4 = 40320 / 4 = 10080. And 7! = 5040. So why is there a factor of 2 difference?

Wait, maybe there's an error in assuming that all orbits have size 4. But for n=4, all orbits have size 4, and it works. For n=8, same reasoning: if no non-identity rotation fixes any arrangement, then all orbits have size 4, so number of orbits is 8! / 4. But why does fixing one person give a different answer?

Wait, maybe fixing one person doesn't account for some symmetries? Or maybe the table has more symmetries?

Wait, the problem says "Two configurations are considered equivalent if one is a rotation of another." So it's only rotations, not reflections. But what about reflections? If we consider reflections as different, then they are not part of the group. So group is only rotations.

Wait, but if we fix one person, how does that relate?

Wait, let me try with n=4 seats, fixing one person. Suppose seat 1 is fixed with person A. Then the remaining 3 people can be arranged in 3! ways. This corresponds to the orbits where person A is at seat 1. Each orbit has 4 arrangements, so total number of orbits is 3! * 4 / 4 = 3! ? No, it's total number of orbits is 3! ? Wait, no. If there are 3! arrangements with person A at seat 1, and each orbit contains 4 arrangements, then the number of orbits is 3! / 4? But that would be 6 / 4, which is not an integer. Contradiction. Wait, this suggests something is wrong.

But in reality, for n=4, we know the number of orbits is 6. If we fix person A at seat 1, we have 3! = 6 arrangements. Each such arrangement is in an orbit of size 4? But that would mean total number of arrangements is 6 * 4 = 24, which is correct. But each orbit is counted 4 times, one for each position of person A. So total number of orbits is 6 / 4 * 4? No, I'm confused again.

Wait, here's the key. Each orbit contains exactly one arrangement with person A at seat 1, one with person A at seat 2, etc. But in reality, for n=4, each orbit contains 4 arrangements, each with person A at different seats. Therefore, the number of orbits is equal to the number of arrangements with person A at seat 1 divided by 1 (since each orbit contributes one such arrangement). Wait, no. If there are M orbits, each orbit contains one arrangement with person A at seat 1, one at seat 2, one at seat 3, one at seat 4. Therefore, total number of arrangements with person A at seat 1 is M. But we know that number is 3! = 6. Hence, M = 6. Which matches the Burnside count of 6. Therefore, number of orbits is 6, and number of arrangements with person A at seat 1 is also 6. Wait, but how is this possible?

If each orbit contains one arrangement with person A at seat 1, then total number of such arrangements is equal to the number of orbits. But we also have that number of such arrangements is 3! = 6. Therefore, number of orbits is 6. Which matches. Therefore, for n=4, it works.

For n=8, similarly, the number of arrangements with person A at seat 1 is 7! = 5040. Each orbit contains exactly one such arrangement. Therefore, number of orbits is 5040. But according to Burnside's lemma, the number of orbits is 8! / 4 = 10080. Contradiction!

Wait, this suggests that my previous assumption that each orbit contains exactly one arrangement with person A at seat 1 is wrong.

Wait, but for n=4, it worked. So what's different with n=8?

Wait, let me think with n=4. If I have an orbit, it contains 4 arrangements, each with person A at different seats. Therefore, the number of orbits is equal to the number of arrangements with person A at seat 1 (which is 6) because each orbit has exactly one arrangement with person A at seat 1. Hence, number of orbits is 6.

But for n=8, if each orbit contains one arrangement with person A at seat 1, then number of orbits is 7! = 5040. But according to Burnside's lemma, it's 8! / 4 = 10080. Therefore, there's a contradiction. Therefore, my assumption that each orbit contains exactly one arrangement with person A at seat 1 must be wrong.

Wait, why? Because in n=4, each orbit has four positions for person A, but in n=8, maybe it has more?

Wait, no. For any orbit, the number of positions where person A can be is equal to the number of rotations. Wait, the orbit size is 4, so for each arrangement in the orbit, person A is in one of the 4 positions. So for each orbit, there are 4 arrangements with person A in different seats. Hence, the number of orbits should be equal to the number of arrangements with person A at seat 1 divided by 1 (since each orbit contains one such arrangement). Wait, no, if each orbit has 4 arrangements with person A in different seats, then the number of orbits is equal to the number of arrangements with person A at seat 1 multiplied by 1 (since each orbit contains one arrangement at seat 1). Wait, this is confusing.

Let me think of it as follows. Total number of arrangements is 8!. Number of orbits is 8! / 4 = 10080. Number of arrangements with person A at seat 1 is 7! = 5040. If each orbit contains 1 arrangement with person A at seat 1, then number of orbits is 5040. But 5040 ≠ 10080. Therefore, each orbit must contain 2 arrangements with person A at seat 1. But how?

Wait, this suggests that my previous assumption that each orbit has size 4 is wrong. If each orbit has size 2, then number of orbits is 8! / 2 = 20160. But this contradicts Burnside's lemma.

Alternatively, maybe some orbits have size 4 and some have smaller size. But earlier, we thought that no non-identity rotations fix any arrangement, so all orbits have size 4. So why the discrepancy?

Wait, let me try to think of a specific example. Let's take two arrangements: one where person A is at seat 1 and person B is at seat 2, and another arrangement where person A is at seat 1 and person B is at seat 3. Are these in the same orbit? Or different orbits?

If you rotate the first arrangement by 90°, person A moves to seat 3, and person B moves to seat 4. If the second arrangement has person A at seat 1 and person B at seat 3, then rotating the first arrangement 90° would give a different arrangement. So these are different orbits.

But if I have an arrangement 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 C is at seat 2, these are in different orbits.

So, if I fix person A at seat 1, each different arrangement is in its own orbit. Therefore, the number of orbits where person A is at seat 1 is equal to the number of such arrangements, which is 7! = 5040. But according to Burnside, total number of orbits is 10080, which is double. So where are the other 5040 orbits?

Ah! Wait, maybe the orbits are not just those where person A is at seat 1, but also where person A is at seat 3, 5, 7. But if we fix person A at seat 1, we are only counting orbits that contain an arrangement with person A at seat 1. But there are other orbits where person A is at seat 3, 5, or 7. However, since we're considering equivalence up to rotation, those orbits are already counted when we fix person A at seat 1. Wait, no. If we fix person A at seat 1, we are only considering arrangements where person A is at seat 1, but the other orbits where person A is not at seat 1 are not considered. Hence, the total number of orbits is equal to the number of orbits where person A is at seat 1, because by rotating, any orbit can have person A at any seat. Wait, but this would mean that the total number of orbits is equal to the number of arrangements with person A at seat 1, which is 7! = 5040. But Burnside says it's 10080.

This is very confusing. Maybe I need to reference a textbook or recall the formula for circular permutations with rotational symmetry.

Wait, in general, the number of distinct circular arrangements of n people is (n-1)! because you fix one person and arrange the rest. But in this case, the table is square, but we are considering rotations as equivalent. If the table were circular, the number would be (8-1)! = 5040. But the problem says it's a square table, so maybe the symmetry group is smaller? Wait, no, the problem says that two configurations are equivalent if one is a rotation of another. So it's still considering rotational symmetry, regardless of the table shape. So if it were a circular table, the number would be (8-1)! = 5040. But since it's a square table, with rotational symmetry of order 4, the number should be 8! / 4 = 10080. But this contradicts the circular table result.

Wait, why? Because for a circular table, the rotational symmetry group has order 8 (rotations by 360/8 = 45°), so number of distinct arrangements is 8! / 8 = 7! = 5040. For a square table, rotational symmetry group has order 4, so number of distinct arrangements is 8! / 4 = 10080. Ah! So this makes sense. The difference is the size of the symmetry group. For a circular table, we have more symmetries (rotations by 45° increments), while for a square table, only rotations by 90° increments.

Therefore, the answer depends on the symmetry group. If the problem considers rotations by any degree (i.e., circular table), the answer is 7! = 5040. If it considers rotational symmetry only to the square's symmetry (rotations by 90°), then the answer is 8! / 4 = 10080.

But the problem says "a square table with 2 people on a side". It doesn't specify whether rotations are considered equivalent only for the square's symmetry or for the full circular symmetry. However, usually, when a table is described as square, it's implied that rotations are considered equivalent only under the square's symmetries. But the problem specifically says "Two configurations are considered equivalent if one is a rotation of another."

But what is a rotation of a square table? A rotation by 90°, 180°, 270°, or 360° (identity). Therefore, the rotational symmetry group is of order 4, hence number of distinct arrangements is 8! / 4 = 10080.

But then why does fixing one person give 7! = 5040? Because fixing one person removes the rotational symmetry. Wait, but if we fix one person, we are effectively choosing a representative from each orbit. But if the orbits have size 4, then the number of orbits is 8! / 4. If I fix one person, I'm counting the number of orbits where that person is at a specific seat, which is 7!, but this would be equal to total number of orbits divided by 4 (since each orbit contains 4 different positions for that person). Hence, total number of orbits is 4 * 7! = 4 * 5040 = 20160, which contradicts Burnside's lemma result of 10080.

Wait, but this suggests that fixing one person is not the correct way to count. Hence, the correct way is Burnside's lemma, giving 10080.

But then why in the circular table case, fixing one person gives the correct result? Because in the circular table, the rotational symmetry group has order 8, so number of orbits is 8! / 8 = 7! = 5040. If you fix one person, you get 7! = 5040, which is the same as 8! / 8. Hence, it works.

Therefore, in the square table case, since the rotational symmetry group has order 4, the number of orbits is 8! / 4 = 10080. Fixing one person would not give the same result, because fixing one person removes the rotational symmetry, but the number of orbits should be equal to the number of arrangements with one person fixed. But this doesn't match.

Wait, maybe the error is in assuming that fixing one person accounts for all symmetries. But if the symmetry group has order 4, fixing one person might not fully account for it. Wait, no. Fixing one person should eliminate all rotational symmetries, as any rotation would move that person to another seat, which is fixed. Hence, fixing one person should give the correct count. But why the discrepancy?

Wait, I think the key is that in the square table case, fixing one person does not fully account for the group action. Let me think of it this way: if I fix one person at a seat, I'm effectively considering arrangements where that seat is fixed, which eliminates the rotational symmetry. Hence, the number of distinct arrangements should be equal to 7!. But according to Burnside's lemma, it's 8! / 4.

Therefore, one of these must be wrong. But in the case of the circular table, both methods agree. So why discrepancy here?

Wait, let's take a small example with n=4, but with a square table (4 seats, one per side). If we use Burnside's lemma, we get (4! + 0 + 0 + 0)/4 = 6. If we fix one person, we get 3! = 6. So it works.

Now, take n=8, but with a square table, and try to apply the same logic. If I fix one person, the number of arrangements is 7! = 5040. But according to Burnside's lemma, it's 8! / 4 = 10080.

Wait, but perhaps in this case, fixing one person is not the correct way because the square table has more symmetries? No, it's only rotations.

Wait, let me think about what fixing one person means. If I fix person A at seat 1, then rotations that move seat 1 are no longer allowed. Hence, the number of arrangements should be 7! = 5040. But according to Burnside's lemma, it's 10080. There must be a miscalculation in the Burnside's lemma part.

Wait, let me go back to the Burnside's lemma. For the square table with 8 seats, group has 4 rotations: 0°, 90°, 180°, 270°. Fixed arrangements:

- 0°: 8! = 40320

- 90°: 0

- 180°: ?

Wait, earlier I thought that fixed arrangements under 180° is 0, but maybe it's not?

Wait, if an arrangement is fixed under 180° rotation, then each person is seated opposite to themselves. But since the table has 8 seats, the opposite seat of seat i is seat i + 4. So for the arrangement to be fixed under 180° rotation, the person at seat i must be the same as person at seat i + 4. But since all people are distinct, this is only possible if each pair of opposite seats has the same person, which is impossible. Hence, fixed arrangements under 180° is 0.

But wait, what if the arrangement is such that person A is at seat 1 and seat 5, but since people are distinct, this is impossible. Hence, fixed arrangements under 180° is 0.

So Burnside's lemma gives (40320 + 0 + 0 + 0)/4 = 10080.

But how can this be? If fixing one person gives 5040, which is half of 10080? Is there a possibility that the problem considers rotations and reflections as equivalent? But the problem says only rotations.

Wait, if we considered reflections as well, the group would have 8 elements (rotations and reflections), and the count would be different. But the problem only mentions rotations.

Wait, maybe the error is that when we fix one person, we are fixing a specific seat, but on a square table, there are symmetries that could still be present. Wait, no. If you fix a person at a specific seat, you eliminate all rotational symmetries.

Wait, let's think about it in terms of the orbit. If we have an arrangement with person A fixed at seat 1, then its orbit includes all rotations of this arrangement. But if person A is fixed at seat 1, then none of the rotations can move person A, so the orbit contains only the arrangement itself. Hence, fixing one person should give the correct count, as each orbit is represented once. Hence, the number of orbits is 7! = 5040.

But this contradicts Burnside's lemma result. Where is the error?

Ah! I think the error is in the application of Burnside's lemma. The group actions are on the seats, but when we fix a seat, we are not considering the group action. Wait, let me recall: Burnside's lemma counts the number of orbits under the group action. In this case, the group is rotations of the table, which act on the set of seatings.

If we fix one person at a seat, we are effectively choosing a representative for each orbit by fixing that person at that seat. Hence, the number of such representatives is equal to the number of orbits.

But if fixing one person gives 7! = 5040, then the number of orbits is 5040, and Burnside's lemma must have been applied incorrectly.

But why did Burnside's lemma give 10080? Because I thought fixed arrangements under rotations were 0, but maybe there are fixed arrangements under 180°?

Wait, let me think again about fixed arrangements under 180° rotation. Suppose we have an arrangement fixed under 180° rotation. This means that the person at seat 1 is the same as the person at seat 5, seat 2 same as seat 6, etc. But since all people are distinct, this is impossible. Therefore, fixed arrangements under 180° is 0.

Wait, but what if the table has an even number of seats and we have pairs? For example, if the table has 4 seats, and we have an arrangement fixed under 180° rotation, it requires pairs of people. But since all people are distinct, this is impossible. Hence, fixed under 180° is 0.

So why is there a discrepancy?

I think the key is that when we fix one person, we are not accounting for the fact that some rotations might have fixed points. But wait, no, we already considered that.

Wait, let me check with n=4 again. Using Burnside's lemma, we get 6. Fixing one person gives 6. So it works. For n=8, why not?

Wait, let's try to calculate the number of orbits using the formula for circular tables. If the table were circular, with 8 seats, the number of arrangements would be 7! = 5040. But if it's a square table, why is it different? Because for a circular table, the group has 8 elements (rotations by 45°, 90°, ..., 315°), while for a square table, the group has 4 elements (rotations by 90°, 180°, 270°, 360°). Therefore, for the square table, the number of distinct arrangements is 8! / 4 = 10080, while for the circular table, it's 7! = 5040.

Therefore, the answer depends on the symmetry group. Since the problem says it's a square table, we should use the square's symmetry group. Hence, the answer is 8! / 4 = 10080.

But why does fixing one person give a different answer? Because fixing one person assumes that the symmetry group is of size 8, but in reality, the symmetry group is of size 4. Hence, fixing one person is not the correct method here.

Wait, but how? If I fix one person, I am removing the rotations. If the symmetry group has size 4, then fixing one person should give 8! / 4 = 10080. But no, fixing one person is different.

Wait, let me think. If the symmetry group has size 4, and we fix one person, how many arrangements are there? It should be total number of arrangements divided by the size of the group. Wait, no.

Actually, the number of orbits is equal to the total number of arrangements divided by the size of the group, if all orbits have the same size. Which is the case here, since no non-identity element fixes any arrangement. Hence, number of orbits is 8! / 4 = 10080.

But if I fix one person, I should get 10080, but I get 7! = 5040. This suggests that fixing one person is not the correct way here.

Wait, but in the circular table example, fixing one person works. Why does it work there? Because in the circular table, fixing one person is equivalent to dividing by the size of the group (which is 8), giving 8! / 8 = 7!.

For the square table, if we fix one person, it's equivalent to dividing by the size of the group (4)? But why does this give 8! / 4 = 10080, but fixing one person gives 7! = 5040? It doesn't.

Wait, no. If you fix one person, you are not dividing by the group size, but fixing a position. How is this related?

Let me think of it as follows: the number of distinct arrangements is equal to the number of ways to arrange the people divided by the number of rotational symmetries, if each orbit has size equal to the number of symmetries. Hence, it's 8! / 4 = 10080.

On the other hand, fixing one person gives 7! = 5040, which is 8! / 8. This suggests that fixing one person assumes a larger symmetry group (8), but in reality, the symmetry group is 4. Hence, fixing one person is not the correct approach here.

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

But why does fixing one person work for the circular table? Because in the circular table, the symmetry group is size 8, and fixing one person is equivalent to dividing by 8, giving 8! / 8 = 7! = 5040, which matches.

For the square table, if we want to use the same logic, we should divide by 4, giving 8! / 4 = 10080.

Therefore, the correct answer is 10080, and the confusion with fixing one person is because fixing one person assumes a different symmetry group.

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

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

To solve the problem of seating 8 people around a square table with **2 people on each side**, and considering that **rotational symmetry makes some configurations equivalent**, we proceed as follows:

---

### **Understanding the Setup**

- The table has **4 sides**, and **2 seats per side**, so **8 total seats** are available.
- We are to place **8 distinct individuals** in these 8 seats.
- **Rotations** that result in the same seating layout are considered **equivalent**: if you rotate the table by 90°, 180°, or 270°, those are not counted as new configurations.

Thus, we are counting **distinct seating arrangements up to rotation**.

---

### **Identifying the Symmetry Group**

The key is to understand which **rotational symmetries** are relevant. A square has **rotational symmetry of order 4**, meaning that rotating the table by 90°, 180°, 270°, or 360° (identity) gives equivalent configurations. So, the group of symmetries (rotations) has **4 elements**.

---

### **Using Burnside's Lemma to Count Distinct Arrangements**

Burnside's Lemma is ideal for this type of problem: it tells us that the number of distinct arrangements (under group action) is the **average number of arrangements fixed** by each group element.

Let’s define the group:
- Identity (0° rotation): Fixes all arrangements.
- Rotation by 90°: No arrangement is fixed (because a 90° rotation maps each seat to a new one, requiring all seats to have the same person, which is impossible since all people are distinct).
- Rotation by 180°: Similarly, no arrangement is fixed (each person is opposite someone else; again, conflict with distinctness).
- Rotation by 270°: Same reasoning as 90°.

So, the only group element that fixes any arrangement is the identity. Therefore:

- Number of arrangements fixed by identity: $ 8! = 40320 $
- Number of arrangements fixed by 90°, 180°, and 270°: 0 each

Now, applying Burnside’s Lemma:

$$
\text{Number of distinct arrangements} = \frac{1 \times 8! + 0 + 0 + 0}{4} = \frac{40320}{4} = \boxed{10080}
$$

---

### **Why Fixing a Person Isn't Directly Applicable Here**

A common technique in circular permutation is to *fix one person* to account for rotation. But this works *only* if the group of symmetries is of size *equal to the number of seats*, which is the standard circular symmetry (rotating 360° / n).

In our case, the group size is **4** (due to the square's symmetry), not 8. Hence, fixing one person and dividing by that number (or directly calculating $7!$) is **not** appropriate here. Doing so introduces a conceptual mismatch with the actual group of symmetries.

Thus, the **correct** and **consistent** method to apply is **Burnside’s Lemma**, which considers the symmetry group of the square table accurately.

---

### **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 $\dfrac{8!}{4} = \boxed{10080}$.