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. The problem says that two configurations are equivalent if one is a rotation of another. Hmm, let's start by visualizing the table. A square has four sides, and each side has two seats. So, in total, there are 4 sides × 2 seats = 8 seats. That makes sense with the 8 people.

First, if the table wasn't circular or had fixed positions, the number of ways to arrange 8 people would just be 8! (which is 40320). But since the table is circular and rotations are considered the same, we need to adjust for that. In circular permutations, we usually divide by the number of positions because rotating the entire arrangement doesn't create a new configuration. But here, it's a square table with fixed sides. Wait, does that change things?

Wait, the problem says "Two configurations are considered equivalent if one is a rotation of another." So, the key here is that rotations of the entire table don't count as different. So, for example, if you rotate the entire table by 90 degrees, that's considered the same arrangement. But since the table is square, there are four rotational symmetries: 0°, 90°, 180°, and 270°. So, does that mean we need to divide by 4 to account for these rotations?

But hold on, maybe not. Because in circular permutations, we fix one position to eliminate rotational symmetry. But here, the table is square, but each side has two seats. So, maybe the rotational symmetries are different. Let me think.

If we fix one seat, say the seat at the top-left corner, then the person sitting there is fixed, and the rest can be arranged relative to that person. But since there are two seats on each side, maybe fixing one seat affects the rotational symmetry. Alternatively, maybe each rotation by 90 degrees would shift each pair of seats. For example, rotating the table 90 degrees would move each side's two seats to the next side. So, if we rotate the entire table, the pairs of seats move to different sides. Therefore, to account for rotational symmetry, we need to divide by 4.

But wait, let me confirm. If we fix the table in space, meaning that the sides are labeled (e.g., North, East, South, West), then rotating the table would change the positions of the seats. However, if the table is considered the same up to rotation, then seating arrangements that can be rotated into each other are equivalent. So, in that case, the number of distinct arrangements would be 8! divided by 4 (the number of rotational symmetries). But is that the case here?

Alternatively, maybe the problem is considering the table as a circular table with 8 seats arranged in a square, but each side has two seats. So, each side is a pair of adjacent seats. Therefore, rotating the entire table by 90 degrees would cycle the sides, but each side still has two seats. So, if we consider the table's rotational symmetry, each rotation by 90 degrees would permute the sides, but the seats on each side are fixed relative to their side. Hmm, this is getting a bit confusing.

Wait, perhaps we need to model this as a circular permutation problem where the table has 8 seats arranged in a square. In circular permutations, the number is (n-1)! because rotations are considered the same. But here, since the table is square, it might have more symmetries. Wait, but the problem says "Two configurations are considered equivalent if one is a rotation of another." So, maybe it's just considering rotational symmetry, not reflective symmetry. So, the group of rotations is of order 4 (0°, 90°, 180°, 270°). Therefore, the number of distinct arrangements would be 8! divided by 4.

But wait, let's test this with a smaller example. Suppose we have 4 seats around a square table, one per corner. Then, the number of distinct seating arrangements would be (4-1)! = 6, but if we consider rotational symmetry, it's 4! /4 = 24 /4 =6. So, that works. So, in the case of 4 seats, the formula is 4! /4 =6. So, similarly, for 8 seats, 8! /4 = 40320 /4 =10080. But wait, but each side has two seats. So, maybe the problem is different because each side has two seats, so maybe the rotational symmetry isn't just 4. Wait, if each side has two seats, then rotating the entire table by 180 degrees would swap the two seats on each side. Hmm, but 180 degrees is half a rotation. Wait, let's think again.

If we have a square table with two seats on each side, then each side has two adjacent seats. If we rotate the table 180 degrees, each seat moves to the seat diagonally opposite. For example, the top-left seat moves to the bottom-right seat, the top-right moves to the bottom-left, etc. But since each side has two seats, the 180-degree rotation would swap the two seats on the top side with the two seats on the bottom side. Similarly, the left side's seats would swap with the right side's seats. Therefore, in this case, the rotational symmetry group is still of order 4. So, 0°, 90°, 180°, 270°. Therefore, the number of distinct arrangements would be 8! /4 =10080.

But wait, let me check with a different approach. Suppose we fix one seat to eliminate rotational symmetry. Let's say we fix the person sitting at the top-left seat. Then, the remaining 7 people can be arranged in 7! ways. However, since the table has sides with two seats, maybe fixing one seat affects the rotational symmetry. Wait, if we fix one seat, then rotating the table by 90 degrees would move the fixed seat to a different position, which is now occupied by another person. Therefore, by fixing one seat, we have eliminated the rotational symmetry. Therefore, the number of arrangements would be 7!.

But wait, this contradicts the previous result. Which one is correct?

Wait, let's take a simpler case. Suppose we have a square table with 4 seats, one per corner. If we fix one seat, say seat A, then the remaining 3 seats can be arranged in 3! ways. So, 6 arrangements. However, considering rotational symmetry, 4! /4 =6, which matches. So, fixing one seat gives us (n-1)! arrangements. Therefore, in the original problem, if we fix one seat, then the number of arrangements is 7!.

But in the original problem, there are 8 seats. If we fix one seat, then the remaining 7 can be arranged in 7! ways. But does fixing one seat account for all rotational symmetries? Because when you fix a seat, you're effectively saying that rotations that move that seat are considered different, but in reality, rotations that move the seat to another seat would result in a different arrangement. Wait, but in the problem statement, two configurations are considered equivalent if one is a rotation of another. So, if we fix one seat, then we are essentially considering that seat as a reference point, and rotations that move that seat are not considered equivalent. Therefore, by fixing one seat, we are effectively removing the rotational symmetry. Therefore, the number of arrangements would be 7!.

But wait, this contradicts the earlier approach where we divided by 4. Which is correct?

Wait, in the case of 4 seats, the formula (n-1)! gives 6, which is correct. So, if we fix one seat, we get 3! =6. Similarly, for 8 seats, fixing one seat would give 7! =5040. But according to dividing by 4, we get 10080. So, which is correct?

Wait, perhaps the problem is considering that the table is square, so that the rotational symmetries are 4. Therefore, the group of rotations is of order 4. So, the number of distinct arrangements should be 8! /4 =10080. But when we fix one seat, we get 7! =5040. These are different results. Therefore, one of the approaches must be wrong.

Wait, maybe fixing one seat doesn't account for all rotational symmetries. Because if you fix one seat, you are not accounting for the possibility that the entire table can be rotated in such a way that the fixed seat remains fixed. Wait, but if you fix a seat, then any rotation that moves that seat would result in a different arrangement. Therefore, by fixing one seat, you are effectively considering that arrangement up to rotation, because rotations that move the fixed seat are considered different. Therefore, fixing one seat would be the correct approach to eliminate rotational symmetry, giving (n-1)!.

But in the case of a square table with 8 seats, if we fix one seat, then the number of arrangements is 7! =5040. However, if we consider that the table can be rotated in 4 ways (0°, 90°, 180°, 270°), then the total number of arrangements would be 8! /4 =10080. So, why the discrepancy?

Wait, perhaps the problem is that when you fix one seat, you are fixing one of the 8 seats, but the symmetries of the table are such that some symmetries (rotations) leave the fixed seat in place. For example, if you fix the top-left seat, then rotating the table 180 degrees would move the top-left seat to the bottom-right seat, which is different. Therefore, fixing one seat doesn't eliminate all rotational symmetries. Therefore, to account for all rotational symmetries, we need to divide by the number of rotational symmetries, which is 4. Therefore, the correct number is 8! /4 =10080.

But then why does fixing a seat give a different answer? Because when you fix a seat, you're effectively removing the rotational symmetry that would otherwise allow the seat to move. However, in the problem statement, two configurations are equivalent if one is a rotation of another. So, if we fix one seat, then we are not considering rotations that move that seat as equivalent. Therefore, in that case, fixing a seat is actually not the right approach, because the problem considers such rotations as equivalent. Therefore, we shouldn't fix a seat. Therefore, the correct approach is to divide by the number of rotational symmetries, which is 4, giving 8! /4 =10080.

Wait, let's test this with a smaller example. Suppose we have a square table with 4 seats, one per corner. If we consider rotational symmetry, the number of distinct arrangements is 4! /4 =6. If we fix one seat, say seat A, then the remaining 3 seats can be arranged in 3! =6 ways, which gives the same result. So, in that case, fixing a seat gives the correct result. But in the case of 8 seats, why is there a discrepancy?

Wait, in the case of 4 seats, each rotation by 90 degrees cycles all four seats. So, fixing one seat, you have 3! arrangements. However, in the case of 8 seats, each rotation by 90 degrees cycles the four sides. Each side has two seats. So, rotating the table by 90 degrees would move each pair of seats to the next side. For example, the top-left seat moves to the top-right seat, which is part of the same side? Wait, no. Wait, each side has two adjacent seats. So, if you rotate the table 90 degrees, each seat moves to the next seat in the clockwise or counter-clockwise direction. Wait, perhaps each seat is adjacent to two others. So, rotating the table 90 degrees would move each seat to the position of the next seat in the clockwise direction. Therefore, each seat is moved to the next seat in the circular arrangement. Therefore, in this case, the rotational symmetry is similar to a circular permutation, but with 8 seats arranged in a square.

Therefore, in this case, the number of rotational symmetries is 4, as in the square can be rotated 0°, 90°, 180°, 270°, and each rotation would cycle the seats. Therefore, the number of distinct arrangements would be 8! divided by 4, which is 10080.

But when we fix one seat, we get 7! =5040. So, why is this different? Because when we fix one seat, we're not considering the rotational symmetry that would move that seat. However, in the problem statement, the equivalence is defined as rotations of the entire table. So, if we fix a seat, then we are not allowing rotations that move that fixed seat, which would otherwise be considered equivalent. Therefore, to account for all rotational symmetries, we need to divide by 4, regardless of whether we fix a seat or not.

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

Wait, but let me think again. Let's say we fix one seat. Then, the number of arrangements is 7! =5040. But if we don't fix a seat, the total number is 8! =40320. However, when considering rotation equivalence, two arrangements are equivalent if one can be rotated to the other. Therefore, the number of equivalence classes is 8! divided by the number of rotational symmetries, which is 4. Therefore, 40320 /4 =10080. So, that's the answer.

But let me check with another approach. Suppose we number the seats from 1 to 8 in a clockwise manner around the table. Then, two seatings are equivalent if one can be rotated to the other. So, seating arrangements are considered the same if each person is rotated by k positions for some k. Therefore, the number of distinct arrangements is 8! divided by 8, which would be 5040. Wait, but the problem says that two configurations are equivalent if one is a rotation of another. So, if you can rotate the entire table by any number of positions (0 to 7), then the number of distinct arrangements would be 8! /8 =5040. But in the problem, the table is square, which can be rotated in 4 distinct ways (0°, 90°, 180°, 270°). So, why is that?

Wait, maybe the key is that the table is square, so that some rotations correspond to the same physical rotation. For example, rotating the table 90 degrees is equivalent to rotating it 270 degrees in the opposite direction. But perhaps in terms of the seat positions, rotating 90 degrees is different from rotating 180 degrees. Wait, no. Each rotation by 90 degrees cycles the seats in a different way. For example, rotating by 90 degrees moves seat 1 to seat 2, seat 2 to seat 3, seat 3 to seat 4, etc. Rotating by 180 degrees moves seat 1 to seat 5, seat 2 to seat 6, etc. So, these are different transformations. Therefore, the rotational symmetries are indeed 4 in total. Therefore, the number of distinct arrangements is 8! /4 =10080.

But then why does the circular permutation formula give 8! /8 =5040? Because in a circular table with no fixed sides, all rotations (including 1°, 2°, etc.) are considered equivalent. But here, the table is square, so only rotations by multiples of 90° are considered equivalent. Therefore, the number of symmetries is 4, not 8.

Therefore, the correct answer is 10080. So, dividing by 4 instead of 8.

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

But let me check once more with a different approach. Suppose we fix one person's seat to eliminate rotational symmetry. Then, the remaining 7 people can be arranged in 7! ways. However, in the square table, fixing one seat might not eliminate all rotational symmetries. For example, if we fix the top-left seat, then rotating the table 180 degrees would move the top-left seat to the bottom-right seat. Therefore, if we fix one seat, we're not accounting for the 180-degree rotation. Therefore, fixing one seat gives 7! arrangements, but that might not account for all symmetries. Therefore, perhaps we need to divide by 4 regardless of fixing a seat.

Alternatively, if we fix one seat, then the number of distinct arrangements is 7! / (number of remaining symmetries). But if we fix one seat, then the remaining symmetries would be the rotations that fix that seat. For example, in a square, the only rotation that fixes a seat is the identity rotation (0°). Because any other rotation would move that seat. Therefore, if we fix one seat, then the number of rotational symmetries is only 1 (the identity). Therefore, the number of distinct arrangements would be 7! /1 =5040. But this contradicts the previous result.

Wait, maybe the confusion arises because when we fix a seat, we are effectively removing the rotational symmetries that would otherwise move that seat. Therefore, if we fix a seat, the number of distinct arrangements is 7! / (number of symmetries that fix that seat). But in the problem, the equivalence is defined by any rotation, not just those fixing a seat. Therefore, perhaps fixing a seat is not the correct approach here.

Therefore, to account for all rotational symmetries (0°, 90°, 180°, 270°), regardless of whether they fix a seat or not, we need to divide the total permutations by 4. Therefore, 8! /4 =10080.

Therefore, the answer should be 10080. But let's check another perspective. Suppose we number the seats from 1 to 8 in a clockwise manner. Then, two seatings are equivalent if one can be obtained by rotating the other by 0°, 90°, 180°, or 270°. Therefore, the number of distinct seatings is 8! divided by 4, which is 10080. Therefore, that's the answer.

Alternatively, if we consider the table as a circular arrangement with 8 seats, but the table has a square structure, then the rotational symmetries are only those that map the table onto itself. Since the table is square, it can be rotated by 0°, 90°, 180°, 270°, and these are the only rotations that leave the table looking the same. Therefore, the size of the symmetry group is 4. Therefore, the number of distinct arrangements is 8! /4 =10080.

Therefore, after considering multiple approaches and verifying with smaller examples, the conclusion is that the number of distinct seating arrangements is 8! divided by 4, which equals 10080.
Therefore, the number of ways 8 people can be seated around a square table with 2 people on a side, considering rotations as equivalent configurations, is $\boxed{10080}$.
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