💡Okay, so I have this problem here with a set of data: a, b, 9, 10, 11. The average is 10, and the variance is 2. I need to find the absolute difference between a and b, which is |a - b|. The options are A: 2, B: 4, C: 8, D: 12. Hmm, let me think about how to approach this.First, I remember that the average of a set of numbers is the sum of the numbers divided by how many there are. So, since the average is 10 and there are 5 numbers in the set, the total sum should be 5 times 10, which is 50. Let me write that down:a + b + 9 + 10 + 11 = 50Simplifying that, 9 + 10 + 11 is 30, so:a + b + 30 = 50Subtracting 30 from both sides gives:a + b = 20Okay, so that's one equation: a + b = 20. That's good, but I need another equation to solve for both a and b. That's where the variance comes in.Variance is a measure of how spread out the numbers are. The formula for variance is the average of the squared differences from the mean. The mean here is 10, so each number minus 10 squared, then averaged.The formula for variance (σ²) is:σ² = [(a - mean)² + (b - mean)² + (9 - mean)² + (10 - mean)² + (11 - mean)²] / 5Given that the variance is 2, so:[(a - 10)² + (b - 10)² + (9 - 10)² + (10 - 10)² + (11 - 10)²] / 5 = 2Let me compute each term:(9 - 10)² = (-1)² = 1(10 - 10)² = 0² = 0(11 - 10)² = 1² = 1So, substituting back in:[(a - 10)² + (b - 10)² + 1 + 0 + 1] / 5 = 2Simplify the constants:[(a - 10)² + (b - 10)² + 2] / 5 = 2Multiply both sides by 5 to eliminate the denominator:(a - 10)² + (b - 10)² + 2 = 10Subtract 2 from both sides:(a - 10)² + (b - 10)² = 8Alright, so now I have two equations:1. a + b = 202. (a - 10)² + (b - 10)² = 8I need to solve these simultaneously to find a and b. Let me think about how to do this.Since a + b = 20, I can express one variable in terms of the other. Let's solve for b:b = 20 - aNow, substitute this into the second equation:(a - 10)² + ( (20 - a) - 10 )² = 8Simplify the second term:(20 - a - 10) = (10 - a)So, the equation becomes:(a - 10)² + (10 - a)² = 8Wait a minute, (10 - a) is the same as -(a - 10), so when squared, it's the same as (a - 10)². Therefore, both terms are equal:(a - 10)² + (a - 10)² = 8Which simplifies to:2*(a - 10)² = 8Divide both sides by 2:(a - 10)² = 4Take the square root of both sides:a - 10 = ±2So, a = 10 + 2 = 12 or a = 10 - 2 = 8Therefore, if a is 12, then from a + b = 20, b would be 8. If a is 8, then b would be 12.So, the two possible pairs are (a, b) = (12, 8) or (8, 12). Either way, the absolute difference |a - b| is |12 - 8| = 4 or |8 - 12| = 4. So, in both cases, |a - b| is 4.Let me double-check my calculations to make sure I didn't make a mistake.First, checking the sum:If a = 12 and b = 8, then 12 + 8 + 9 + 10 + 11 = 50, which is correct because 50 / 5 = 10, the average.Now, checking the variance:Compute each (x - mean)²:(12 - 10)² = 4(8 - 10)² = 4(9 - 10)² = 1(10 - 10)² = 0(11 - 10)² = 1Sum these up: 4 + 4 + 1 + 0 + 1 = 10Variance is 10 / 5 = 2, which matches the given variance.So, everything checks out. Therefore, |a - b| is 4.**Final Answer**boxed{B}

💡Okay, so I have this geometry problem here about a quadrilateral ABCD. It says that AD is parallel to BC, and ABCD is a parallelogram. Then it asks which of the given conditions must also be satisfied. The options are about the sums of certain angles being 180 degrees.First, I need to recall what defines a parallelogram. From what I remember, a parallelogram is a quadrilateral with both pairs of opposite sides parallel. So, if AD is parallel to BC, then AB must be parallel to CD for it to be a parallelogram. That makes sense because in a parallelogram, opposite sides are not only parallel but also equal in length.Now, the problem mentions that AD is parallel to BC, and ABCD is a parallelogram. So, I guess the key here is to figure out what additional condition must be true based on these facts.Looking at the options, they all involve the sum of two angles being 180 degrees. I remember that in a parallelogram, opposite angles are equal, and consecutive angles are supplementary. That means if you have one angle, the angle next to it will add up to 180 degrees.Let me write down the properties to keep track:1. Opposite sides are parallel: AD || BC and AB || CD.2. Opposite angles are equal: angle A = angle C and angle B = angle D.3. Consecutive angles are supplementary: angle A + angle B = 180°, angle B + angle C = 180°, and so on.Given that, let's look at the options:A: angle A + angle B = 180°B: angle A + angle C = 180°C: angle B + angle C = 180°D: angle B + angle D = 180°From the properties, I know that consecutive angles are supplementary. So, angle A + angle B should be 180°, angle B + angle C should be 180°, angle C + angle D should be 180°, and angle D + angle A should be 180°. But the question is asking which condition must also be satisfied given that AD is parallel to BC and ABCD is a parallelogram. So, I think it's about confirming which of these angle sums must hold true based on the given information.Option A: angle A + angle B = 180°. That's a property of a parallelogram, but is it something that needs to be satisfied in addition to AD being parallel to BC? Since ABCD is already a parallelogram, this condition is already met.Option B: angle A + angle C = 180°. Hmm, in a parallelogram, opposite angles are equal, so angle A = angle C. If angle A + angle C = 180°, that would mean each angle is 90°, making it a rectangle. But the problem doesn't specify that ABCD is a rectangle, just a parallelogram. So this condition isn't necessarily true.Option C: angle B + angle C = 180°. Again, this is a property of a parallelogram because consecutive angles are supplementary. Since ABCD is a parallelogram, this must be true.Option D: angle B + angle D = 180°. In a parallelogram, opposite angles are equal, so angle B = angle D. If angle B + angle D = 180°, then each would be 90°, making it a rectangle. But since ABCD is just a parallelogram, this isn't necessarily true.Wait, but I'm a bit confused here. If ABCD is a parallelogram, then by definition, consecutive angles are supplementary. So, angle B + angle C should be 180°, which is option C. But I also know that in a parallelogram, opposite angles are equal, so angle A = angle C and angle B = angle D. Let me think again. The problem states that AD is parallel to BC, and ABCD is a parallelogram. So, given that AD || BC, and ABCD is a parallelogram, what else must be true? Since ABCD is a parallelogram, we already know that AB || CD. So, with both pairs of opposite sides parallel, the properties of a parallelogram apply.Therefore, the conditions about the angles being supplementary are already satisfied. But the question is asking which condition must also be satisfied. So, maybe it's about confirming that the sides being parallel lead to the angles being supplementary.Wait, perhaps I'm overcomplicating it. Since ABCD is a parallelogram, it must satisfy all the properties, including the consecutive angles being supplementary. So, angle B + angle C = 180° is a necessary condition for it to be a parallelogram.But then why are the other options not correct? Let me check:Option A: angle A + angle B = 180°. This is also a property of a parallelogram, but it's not directly related to the given condition that AD || BC. It's more about the sides AB and CD being parallel.Option B: angle A + angle C = 180°. As I thought earlier, this would imply that angle A and angle C are supplementary, but since they are opposite angles in a parallelogram, they are equal. So, this would only be true if both angles are 90°, which isn't necessarily the case.Option C: angle B + angle C = 180°. This is directly related to the sides AD || BC and AB || CD. Since AD || BC, the consecutive angles along those sides must be supplementary. So, angle B and angle C are consecutive angles along side BC and CD, which are parallel. Therefore, their sum must be 180°.Option D: angle B + angle D = 180°. Again, angle B and angle D are opposite angles in a parallelogram, so they are equal. Their sum being 180° would mean each is 90°, making it a rectangle, which isn't necessarily the case.So, putting it all together, the condition that must be satisfied is that angle B + angle C = 180°, which is option C.

💡Alright, so I have this problem about a 100x100 grid where each small square is painted one of four colors. The condition is that each row and each column contains exactly 25 squares of each color. I need to prove that there exist two rows and two columns such that the four intersecting squares are each a different color. Hmm, okay, let's break this down.First, let me understand the setup. We have a grid with 100 rows and 100 columns. Each row has 25 squares of each color, say A, B, C, D. Similarly, each column also has exactly 25 squares of each color. So, it's a balanced grid in terms of color distribution.I need to find two rows and two columns such that when they intersect, the four squares formed are all different colors. That is, in the 2x2 subgrid formed by these two rows and two columns, each cell has a unique color.Let me think about how to approach this. Maybe I can use the Pigeonhole Principle or some combinatorial argument. Since the grid is large, 100x100, and each row and column has a fixed number of each color, perhaps there's a way to count the number of possible color pairs and show that some configuration must exist.Let me try to count the number of color pairs in a single row. In each row, there are 100 squares, each colored A, B, C, or D, with exactly 25 of each color. So, in a single row, how many pairs of different colors can I have?Well, for two distinct colors, say A and B, the number of pairs would be 25*25, since there are 25 A's and 25 B's. Similarly, for A and C, it's 25*25, and so on for all combinations of two colors. There are C(4,2) = 6 such pairs.So, in one row, the number of color pairs is 6*(25*25) = 6*625 = 3750. That's a lot of color pairs in one row.Now, since there are 100 rows, the total number of color pairs across all rows is 100*3750 = 375,000.But wait, these color pairs are distributed across the columns. Each color pair in a row corresponds to two columns. So, for each row, the 3750 color pairs are spread out over C(100,2) = 4950 possible column pairs.So, on average, each column pair would have 375,000 / 4950 ≈ 75.76 color pairs. Since we can't have a fraction of a color pair, some column pairs must have at least 76 color pairs.This is where the Pigeonhole Principle comes into play. If we have more color pairs than the number of column pairs, some column pairs must share multiple color pairs. Specifically, there must be at least one column pair that has at least 76 color pairs.But what does that mean? If two columns have 76 color pairs, that means there are 76 different rows where these two columns have different colors. So, in these two columns, across 76 rows, each row has a different color combination.Now, let's focus on these two columns and the 76 rows. Each of these rows has a pair of colors in these two columns. Since there are four colors, the possible color pairs are (A,B), (A,C), (A,D), (B,C), (B,D), and (C,D). That's six possible pairs.If we have 76 rows and only six possible color pairs, by the Pigeonhole Principle again, at least one color pair must appear multiple times. Specifically, at least one color pair must appear at least ceiling(76/6) = 13 times.Wait, but if a color pair appears 13 times, that means there are 13 rows where these two columns have the same color pair. But hold on, each column can only have 25 of each color. So, if in one column, say column X, we have 13 rows where the color is A, and in column Y, we have 13 rows where the color is B, that's fine because 13 is less than 25.But the problem is that we need four different colors in the four intersections. So, if we have two rows and two columns, we need all four combinations to be different. If in these two columns, we have multiple rows with the same color pair, that might not help us directly.Maybe I need to think differently. Instead of focusing on the color pairs, perhaps I should look at the distribution of colors in the columns.Each column has 25 of each color. So, if I pick two columns, each has 25 A's, 25 B's, 25 C's, and 25 D's. The intersection of these two columns with any two rows should give us four squares. We want all four to be different colors.So, perhaps I can argue that if I have enough rows, some two rows must interact with the two columns in such a way that all four colors are present.Alternatively, maybe I can use double counting or some combinatorial design.Let me consider the number of possible 2x2 subgrids. There are C(100,2) * C(100,2) possible 2x2 subgrids, which is a huge number. But we need to show that at least one of them has all four colors.But that approach might not be straightforward. Maybe instead, I can fix two columns and analyze the rows.Suppose I fix two columns, say column X and column Y. Each has 25 of each color. Now, for each row, the pair (color in X, color in Y) can be one of 16 possible combinations (since each column has four colors). However, since each color appears exactly 25 times in each column, the number of times each color pair appears is limited.Wait, actually, for two columns, the number of possible color pairs is 4*4=16. But since each color appears 25 times in each column, the number of times a specific color pair appears is at most 25, because you can't have more than 25 of the same color in a column.But actually, it's possible that a color pair could appear up to 25 times if all the positions of a color in column X align with all the positions of another color in column Y. But given that each color appears 25 times, the maximum number of times a specific color pair can appear is 25.But in reality, since the grid is balanced, the distribution might be more even.Wait, maybe I can use an averaging argument. If I fix two columns, how many different color pairs do they have across all rows?Each column has 25 of each color, so the number of color pairs is 4*4=16, but the total number of pairs is 100. So, on average, each color pair appears 100/16=6.25 times.But this is an average, so some color pairs must appear more than 6 times, and some less.But I'm not sure if this helps directly. Maybe I need to consider the number of color pairs across all pairs of columns.Earlier, I calculated that there are 375,000 color pairs across all rows, and these are distributed over 4950 column pairs. So, on average, each column pair has about 75.76 color pairs.But since we can't have a fraction, some column pairs must have at least 76 color pairs.Now, if a column pair has 76 color pairs, that means there are 76 rows where these two columns have different colors. So, in these two columns, across 76 rows, each row has a different color combination.But since there are only 6 possible color pairs (since we're considering two columns, each with four colors, but we're only counting pairs where the colors are different), wait, no, actually, for two columns, the number of possible color pairs is 4*4=16, but we're only considering pairs where the colors are different, which is 12.Wait, no, actually, in the earlier step, I considered color pairs as two different colors in the same row, so for two columns, the color pairs would be (A,B), (A,C), (A,D), (B,A), (B,C), (B,D), etc. But actually, in the context of two columns, the color pairs are ordered, so it's 4*3=12 possible ordered pairs.But in the initial count, I considered unordered pairs, so maybe it's 6 unordered pairs.Wait, I think I need to clarify this.When I counted the number of color pairs in a row, I considered unordered pairs, so for two colors, it's C(4,2)=6. So, in the context of two columns, the number of unordered color pairs is also 6.But in reality, for two columns, the color pairs can be ordered, so it's 12. But since we're dealing with unordered pairs, it's 6.But I'm getting confused here. Let me try to clarify.In a single row, when I count the number of color pairs, I'm considering unordered pairs of two different colors. So, for example, in a row, if I have color A in column X and color B in column Y, that's one unordered pair {A,B}. Similarly, if I have color B in column X and color A in column Y, that's the same unordered pair {A,B}.But in reality, when considering two columns, the ordered pairs matter because the position in the column affects the color. So, maybe I need to consider ordered pairs.Wait, perhaps I should think of it as, for two columns, each row contributes an ordered pair (color in column X, color in column Y). So, for each row, it's an ordered pair, and there are 4*4=16 possible ordered pairs.But since each column has exactly 25 of each color, the number of times each ordered pair appears is limited.Specifically, for each ordered pair (A,B), the number of rows where column X is A and column Y is B is at most 25, because column X has only 25 A's.Similarly, for (A,C), it's at most 25, and so on.But the total number of ordered pairs across all rows is 100, since there are 100 rows.So, if we have 16 possible ordered pairs, and 100 rows, the average number of times each ordered pair appears is 100/16=6.25.But since we can't have fractions, some ordered pairs must appear at least 7 times.Wait, but this is for a specific pair of columns. So, if I fix two columns, the number of times each ordered pair appears is at most 25, but on average, it's 6.25.But I'm not sure if this helps directly. Maybe I need to consider the number of ordered pairs across all column pairs.Wait, earlier I calculated that there are 375,000 color pairs across all rows, and these are distributed over 4950 column pairs. So, on average, each column pair has about 75.76 color pairs.But since each column pair can have at most 25 of each ordered pair, and there are 16 ordered pairs, the maximum number of color pairs for a column pair is 16*25=400, which is way higher than 75.76.Wait, that doesn't make sense. Maybe I'm misunderstanding something.Wait, no, actually, for a specific column pair, the number of color pairs is the number of rows where the two columns have different colors. Since each column has 25 of each color, the number of rows where column X is A and column Y is B is at most 25, as column X has only 25 A's.Similarly, for each of the 12 ordered pairs where the colors are different, the number of rows is at most 25.So, the total number of color pairs for a column pair is at most 12*25=300.But earlier, I calculated that on average, each column pair has about 75.76 color pairs. So, 75.76 is much less than 300, which means that the average is much lower than the maximum possible.But I need to find a column pair that has a certain number of color pairs, such that among those, there must be two rows where the four intersections are all different colors.Wait, maybe I need to think about it differently. Suppose I have two columns, and I look at the colors in those columns across all rows. Each column has 25 of each color, so in total, across 100 rows, each column has 25 A's, 25 B's, 25 C's, and 25 D's.Now, if I consider the pairs of colors in these two columns, there are 16 possible ordered pairs. The total number of ordered pairs across all rows is 100.If I can show that among these 100 ordered pairs, there must be two rows where the four intersections are all different colors, then I'm done.But how?Wait, perhaps I can use the Pigeonhole Principle again. If I have 100 ordered pairs, and only 16 possible types, then some ordered pairs must repeat.But I need to find two rows where the four intersections are all different colors. That is, for two rows r1 and r2, and two columns c1 and c2, the colors at (r1,c1), (r1,c2), (r2,c1), and (r2,c2) are all different.So, in terms of ordered pairs, for rows r1 and r2, the pairs (c1,c2) for r1 and r2 must be such that all four colors are present.Wait, maybe I can think of it as, for two rows, the two ordered pairs must form a Latin square of order 2, meaning all four elements are distinct.So, perhaps I can count the number of such possible combinations and show that at least one must exist.Alternatively, maybe I can use an argument based on the number of possible colorings and show that it's impossible to avoid having such a 2x2 subgrid.Wait, another approach: consider the number of possible colorings for two rows and two columns. There are 4^4=256 possible colorings. But we need only those colorings where all four colors are different, which is 4! = 24.So, the probability that a random 2x2 subgrid has all four colors is 24/256 = 3/32.But since we have a lot of 2x2 subgrids, maybe we can use expectation to show that at least one exists.But I'm not sure if that's the right approach, because the colorings are not random; they're constrained by the row and column sums.Wait, maybe I can use double counting. Let me define a 2x2 subgrid as four cells at the intersection of two rows and two columns. I need to show that at least one such subgrid has all four colors.Let me count the number of such subgrids and the number of colorings that satisfy the condition.But I'm not sure how to proceed with that.Wait, perhaps I can fix two rows and count the number of column pairs that form a 2x2 subgrid with all four colors.If I can show that for some two rows, there are enough column pairs that satisfy the condition, then I'm done.Alternatively, maybe I can use the fact that each column has 25 of each color, so for any two columns, the number of rows where they have the same color is limited.Wait, for two columns, the number of rows where they have the same color is at most 25*4=100, but that's the total number of rows, so that doesn't help.Wait, no, actually, for two columns, the number of rows where they have the same color is the sum over each color of the number of rows where both columns have that color.Since each column has 25 of each color, the number of rows where both columns have color A is at most 25, because column X has only 25 A's, and column Y has only 25 A's. So, the maximum number of rows where both columns have color A is 25.Similarly, for colors B, C, and D, the maximum number of rows where both columns have the same color is 25 for each color.Therefore, the total number of rows where two columns have the same color is at most 4*25=100, which is the total number of rows. But that's just an upper bound; in reality, it's much less because the overlaps can't all be 25.Wait, actually, the number of rows where two columns have the same color is equal to the sum over each color of the number of rows where both columns have that color.Let me denote the number of rows where both columns have color A as x_A, similarly x_B, x_C, x_D. Then, x_A + x_B + x_C + x_D = number of rows where the two columns have the same color.But each x_color is at most 25, because each column has only 25 of that color.But the total number of rows is 100, so the number of rows where the two columns have different colors is 100 - (x_A + x_B + x_C + x_D).But I don't know if this helps directly.Wait, maybe I can use the fact that the number of rows where two columns have different colors is 100 - (x_A + x_B + x_C + x_D). Since each x_color is at most 25, the sum x_A + x_B + x_C + x_D is at most 100, which is the total number of rows. So, the number of rows where the two columns have different colors is at least 0, which is trivial.But I need to find two columns where the number of rows with different colors is large enough to guarantee that among them, there are two rows that form a 2x2 subgrid with all four colors.Wait, maybe I can use the following approach:1. For each pair of columns, count the number of rows where they have different colors. Let's call this number D.2. For each such pair of columns, if D is large enough, then among those D rows, there must be two rows where the four intersections are all different colors.3. Therefore, if we can show that for some pair of columns, D is large enough, then we're done.So, how large does D need to be to guarantee that among those D rows, there are two rows forming a 2x2 subgrid with all four colors?Let me think about it. For two columns, each row contributes an ordered pair of colors. We need two rows such that the four colors are all different.So, in terms of ordered pairs, we need two ordered pairs (a,b) and (c,d) such that {a,b,c,d} = {A,B,C,D}.So, how many such pairs do we need to ensure that this happens?I think this is similar to the problem of finding two edges in a graph that form a certain structure.Wait, maybe I can model this as a graph where each node represents a color, and each edge represents an ordered pair. Then, finding two edges that form a complete graph on four nodes.But I'm not sure if that's helpful.Alternatively, maybe I can think in terms of Ramsey Theory, where we're trying to find a monochromatic or specific substructure.But perhaps a simpler approach is to consider the number of possible color pairs and use the Pigeonhole Principle.If we have D rows where two columns have different colors, then each row contributes an ordered pair of colors. There are 12 possible ordered pairs (since we exclude the same color pairs).If D is large enough, then some ordered pairs must repeat, but I need to ensure that among these, there are two ordered pairs that together form four distinct colors.Wait, let's think about it. Suppose I have D rows, each with an ordered pair of colors from the 12 possible. I need to find two rows such that their ordered pairs together use all four colors.So, for example, if one row has (A,B) and another has (C,D), then together they use all four colors.Similarly, if one row has (A,C) and another has (B,D), that also works.But if all the ordered pairs are, say, (A,B), then we can't form four distinct colors.So, to guarantee that there are two rows with ordered pairs that together use all four colors, we need to ensure that the number of ordered pairs is large enough that it's impossible for all of them to be confined to fewer than four colors.Wait, but each ordered pair uses two colors, so if we have enough ordered pairs, they must cover all four colors.But I'm not sure how to quantify this.Wait, maybe I can use the fact that if we have more than a certain number of ordered pairs, then they must cover all four colors.But I need to find a lower bound on D such that any D ordered pairs must include two pairs that together cover all four colors.Alternatively, perhaps I can use the concept of covering.If I have D ordered pairs, each using two colors, then the total number of color occurrences is 2D.Since there are four colors, the average number of times each color appears is 2D/4 = D/2.If D is large enough, then each color must appear a certain number of times.Wait, but I'm not sure if that helps.Wait, maybe I can think of it as a hypergraph problem, where each ordered pair is a hyperedge connecting two colors, and we need to find two hyperedges that together cover all four colors.But this might be overcomplicating.Wait, perhaps a simpler approach is to consider that for two columns, the number of rows where they have different colors is D. Each such row contributes an ordered pair of colors.If D is greater than the number of possible ordered pairs that don't cover all four colors, then we must have at least one ordered pair that covers all four colors.Wait, but that's not quite right, because each ordered pair only covers two colors.Wait, actually, no. To cover all four colors, we need two ordered pairs that together cover all four colors.So, if we have enough ordered pairs, we must have two that together cover all four colors.So, how many ordered pairs do we need to ensure that?Let me think about the maximum number of ordered pairs that can be formed without having two pairs that together cover all four colors.In other words, what's the maximum number of ordered pairs we can have such that any two pairs share at least one color.This is similar to the concept of intersecting families in combinatorics.In our case, the ordered pairs are from a set of four elements, and we want the maximum number of pairs such that any two pairs share at least one color.This is known as an intersecting family.For a set of size 4, the maximum intersecting family of 2-element subsets is 6, which is all the pairs containing a fixed element.But in our case, the ordered pairs are from a 4-element set, and we want the maximum number of ordered pairs such that any two share at least one color.Wait, actually, for ordered pairs, the situation is a bit different.If we fix a color, say A, then all ordered pairs that include A will share A with each other. So, the number of ordered pairs that include A is 2*3=6 (since for ordered pairs, it's (A,B), (A,C), (A,D), (B,A), (C,A), (D,A)).Similarly, for any fixed color, there are 6 ordered pairs that include it.So, if we take all ordered pairs that include A, we have 6 ordered pairs, and any two of them share A.Similarly, if we take all ordered pairs that include B, we have another 6, and so on.But if we take ordered pairs from different fixed colors, they might not share a common color.Wait, but if we take all ordered pairs that include either A or B, then some pairs will share A or B, but others might not.Wait, actually, if we take all ordered pairs that include A or B, then any two pairs will share A or B or both.But I'm not sure if that's the case.Wait, for example, (A,C) and (B,D) don't share A or B, so they don't share a common color.So, actually, if we take ordered pairs that include A or B, we can still have pairs that don't share a common color.Therefore, the maximum intersecting family for ordered pairs on a 4-element set is not straightforward.Wait, maybe I need to think differently. Let's consider that for ordered pairs, the maximum size of an intersecting family is 12. Because for each color, there are 6 ordered pairs that include it, and if we take all ordered pairs that include A or B, that's 6 + 6 = 12, but some overlap.Wait, actually, the total number of ordered pairs is 16, and the maximum intersecting family is 12, because you can take all ordered pairs that include A or B, which is 12.But I'm not entirely sure about this.Wait, let me check. For a set of size n, the maximum size of an intersecting family of k-element subsets is C(n-1,k-1). But for ordered pairs, it's different.Wait, actually, for ordered pairs, the maximum intersecting family is n(n-1), which for n=4 is 12. Because you can fix one element and take all ordered pairs that include it, which is 2*(n-1) = 6, but if you allow pairs that include either of two fixed elements, it's 2*2*(n-1) = 12.Wait, no, that doesn't make sense.Wait, actually, for ordered pairs, the maximum intersecting family is n(n-1), which is 12 for n=4.But I'm not sure if that's correct.Wait, let's think about it. For ordered pairs, an intersecting family is a set of pairs where any two pairs share at least one element.So, for n=4, what's the maximum size of such a family?One way to construct a large intersecting family is to fix an element, say A, and take all ordered pairs that include A. There are 2*(n-1) = 6 such pairs: (A,B), (A,C), (A,D), (B,A), (C,A), (D,A).Similarly, if we fix two elements, say A and B, and take all ordered pairs that include A or B, that would be 6 + 6 - 2 = 10 pairs, because (A,B) and (B,A) are counted twice.Wait, but actually, if we fix two elements, A and B, the number of ordered pairs that include A or B is:- Pairs where the first element is A: 3 (A,B), (A,C), (A,D)- Pairs where the second element is A: 3 (B,A), (C,A), (D,A)- Similarly for B: 3 (B,A), (B,C), (B,D) and 3 (A,B), (C,B), (D,B)But this counts some pairs twice, like (A,B) and (B,A).So, the total number is 3 + 3 + 3 + 3 - 2 = 10, because (A,B) and (B,A) are counted twice.But actually, the total number of ordered pairs that include A or B is:- All ordered pairs where first element is A or B: 2*4 = 8- All ordered pairs where second element is A or B: 2*4 = 8But this counts the pairs where both elements are A or B twice. There are 2*2=4 such pairs: (A,A), (A,B), (B,A), (B,B).So, using inclusion-exclusion, the total number is 8 + 8 - 4 = 12.Therefore, the maximum size of an intersecting family for ordered pairs on a 4-element set is 12.Wait, but in our case, we're only considering ordered pairs where the two elements are different, because we're excluding same-color pairs.So, the total number of ordered pairs is 12 (since 4*3=12).Wait, no, actually, in our case, the ordered pairs are from two columns, each with four colors, so the total number of ordered pairs is 16, including same-color pairs.But earlier, I was considering only different colors, which are 12 ordered pairs.Wait, I'm getting confused again.Let me clarify:- For two columns, each with four colors, the total number of ordered pairs is 4*4=16.- Out of these, 4 are same-color pairs: (A,A), (B,B), (C,C), (D,D).- The remaining 12 are different-color pairs.So, when I'm considering the number of rows where two columns have different colors, I'm looking at the 12 different-color ordered pairs.Now, if I have D rows where two columns have different colors, then each row corresponds to one of these 12 ordered pairs.If D is large enough, then among these D ordered pairs, there must be two that together cover all four colors.So, what's the maximum number of ordered pairs we can have without having two that together cover all four colors?In other words, what's the maximum number of ordered pairs we can have such that any two pairs share at least one color.This is similar to an intersecting family, but for ordered pairs.From earlier, I think the maximum size of such a family is 12, but that's for all ordered pairs, including same-color ones.Wait, no, in our case, we're only considering different-color ordered pairs, which are 12.So, what's the maximum number of different-color ordered pairs we can have such that any two share at least one color.This is equivalent to finding the largest intersecting family within the 12 different-color ordered pairs.I think this is a known problem in combinatorics.For a set of size n, the maximum intersecting family of 2-element subsets is C(n-1,1) = n-1. But for ordered pairs, it's different.Wait, actually, for ordered pairs, the maximum intersecting family is n(n-1)/2, but I'm not sure.Wait, let's think about it. For ordered pairs, an intersecting family is a set of pairs where any two pairs share at least one element.In our case, the elements are colors, and the pairs are ordered.So, for four colors, how large can such a family be?One way to construct a large intersecting family is to fix a color, say A, and take all ordered pairs that include A.There are 6 such ordered pairs: (A,B), (A,C), (A,D), (B,A), (C,A), (D,A).Similarly, if we fix two colors, say A and B, and take all ordered pairs that include A or B, we get more pairs.But we have to ensure that any two pairs share at least one color.Wait, if we take all ordered pairs that include A or B, then any two pairs will share A or B or both.But actually, no. For example, (A,C) and (B,D) don't share A or B.Wait, so that doesn't work.Wait, actually, if we fix a color, say A, and take all ordered pairs that include A, then any two pairs will share A.Similarly, if we fix two colors, say A and B, and take all ordered pairs that include A or B, but ensuring that any two pairs share at least one color.But as I saw earlier, this isn't possible because pairs like (A,C) and (B,D) don't share a color.Therefore, the maximum intersecting family for ordered pairs on four elements is actually 6, which is all ordered pairs that include a fixed color.So, if we fix color A, we can have 6 ordered pairs: (A,B), (A,C), (A,D), (B,A), (C,A), (D,A).Similarly, for color B, it's another 6, but overlapping with the first set.Therefore, the maximum size of an intersecting family for ordered pairs on four elements is 6.Therefore, if we have more than 6 ordered pairs, we must have two pairs that don't share a common color, meaning they together cover all four colors.Wait, no, actually, if we have 7 ordered pairs, since the maximum intersecting family is 6, then at least one pair must be outside the family, meaning it doesn't share a common color with all others.But I'm not sure if that directly implies that two pairs together cover all four colors.Wait, perhaps I need to think differently.If we have D ordered pairs, and D > 6, then by the Pigeonhole Principle, at least two of them must not share a common color, meaning they together cover all four colors.Wait, let me test this.Suppose we have 7 ordered pairs. If all of them include color A, then they all share A, and we can't form two pairs that together cover all four colors.But if we have 7 ordered pairs, and not all include A, then some must include other colors.Wait, but the maximum number of ordered pairs that include A is 6. So, if we have 7 ordered pairs, at least one must not include A.Similarly, if we have 7 ordered pairs, and none include A, then they must include other colors.Wait, but I'm not sure.Wait, actually, the maximum number of ordered pairs that include a fixed color is 6. So, if we have 7 ordered pairs, at least one must not include that fixed color.But that doesn't necessarily mean that two of them together cover all four colors.Wait, perhaps I need to consider that if we have more than 6 ordered pairs, then we must have two pairs that together cover all four colors.Wait, let's see. Suppose we have 7 ordered pairs. If all 7 include color A, then they all share A, and we can't form two pairs that together cover all four colors.But if we have 7 ordered pairs, and not all include A, then some must include other colors.Wait, but the maximum number of ordered pairs that include A is 6, so if we have 7, at least one must not include A.Similarly, if we have 7 ordered pairs, and none include A, then they must include B, C, or D.But I'm not sure how to proceed.Wait, maybe I can use the following argument:If we have D ordered pairs, and D > 6, then there must be two ordered pairs that together cover all four colors.Because if we have 7 ordered pairs, and each ordered pair uses two colors, then the total number of color occurrences is 14.Since there are four colors, by the Pigeonhole Principle, at least one color must appear at least ceiling(14/4)=4 times.But I'm not sure if that helps.Wait, actually, if a color appears 4 times, then there are four ordered pairs that include that color. But that doesn't necessarily mean that two of them together cover all four colors.Wait, perhaps I need to think about it differently.Suppose we have D ordered pairs. Each ordered pair uses two colors. We need to find two ordered pairs such that their union is all four colors.So, the question is, what's the minimum D such that any D ordered pairs must contain two pairs whose union is all four colors.This is similar to the concept of covering radius in combinatorics.I think that if D is greater than 6, then it's possible to have two pairs that together cover all four colors.Wait, actually, let's consider the complement. What's the maximum number of ordered pairs we can have without having two pairs that together cover all four colors.This is equivalent to the maximum number of ordered pairs such that every pair shares at least one color with every other pair.Wait, no, that's not exactly right. It's the maximum number of ordered pairs such that no two pairs together cover all four colors.Which means that for any two pairs, their union is at most three colors.So, how large can such a family be?I think this is similar to a family of sets where the union of any two sets is not the entire set.In our case, the sets are the colors used in the ordered pairs.So, for four colors, the maximum size of such a family is 6.Because if we fix a color, say A, and take all ordered pairs that include A, then any two pairs will share A, so their union will be at most three colors (A plus two others). Therefore, the union won't cover all four colors.Similarly, if we fix two colors, say A and B, and take all ordered pairs that include A or B, but ensuring that any two pairs share at least one color.Wait, but earlier I saw that this isn't possible because pairs like (A,C) and (B,D) don't share a color.Therefore, the maximum size of such a family is actually 6.Therefore, if we have more than 6 ordered pairs, we must have two pairs that together cover all four colors.Therefore, if D > 6, then there exist two rows such that the four intersections are all different colors.So, going back to our problem, for each pair of columns, the number of rows where they have different colors is D.If D > 6, then we can find two rows that form a 2x2 subgrid with all four colors.But earlier, I calculated that the average D across all column pairs is about 75.76, which is much larger than 6.Therefore, there must exist at least one pair of columns where D > 6, and hence, we can find two rows that form the desired 2x2 subgrid.Wait, but actually, the average D is 75.76, which is much larger than 6, so certainly, there exist column pairs with D > 6.In fact, since the average is 75.76, there must be column pairs with D as high as, well, potentially up to 100, but certainly much higher than 6.Therefore, for such column pairs, we can find two rows that form a 2x2 subgrid with all four colors.Therefore, the proof is complete.So, to summarize:1. Each row has 3750 color pairs.2. Across all rows, there are 375,000 color pairs.3. These are distributed over 4950 column pairs, averaging about 75.76 per column pair.4. Therefore, some column pairs have D > 6.5. For such column pairs, with D > 6, there must be two rows that form a 2x2 subgrid with all four colors.Hence, such a 2x2 subgrid exists.**Final Answer**boxed{4}

💡Okay, so I have this problem where six people are sitting around a round table. I need to figure out the number of possible ordered pairs (f, m), where f is the number of people sitting next to at least one female, and m is the number of people sitting next to at least one male. Hmm, let me try to break this down.First, I think I should consider different configurations of men and women around the table. Since it's a round table, the arrangement is circular, so rotations don't count as different arrangements. That might simplify things a bit.Let me start by considering the extreme cases. What if all six people are men? In that case, there are no women, so f would be 0 because no one is sitting next to a female. On the other hand, m would be 6 because everyone is sitting next to at least one male (in fact, everyone is sitting next to two males). So that gives me the pair (0,6).Similarly, if all six people are women, then f would be 6 because everyone is sitting next to at least one female, and m would be 0 because there are no males. So that gives me the pair (6,0).Now, let's consider cases where there's a mix of men and women. Let's start with one woman and five men. The woman will be sitting between two men. So, the two men next to her will each be sitting next to at least one female. Therefore, f would be 2 because only those two men are next to the woman. As for m, since there are five men, each man is sitting next to at least one male (in fact, all except the two next to the woman are sitting next to two males). So m would still be 6 because everyone is sitting next to at least one male. So that gives me the pair (2,6).Next, let's consider two women. There are two possibilities here: the two women can be sitting together or apart. If they are sitting together, then they form a block. Each woman is sitting next to a man on one side and a woman on the other. So, the two men adjacent to the block of women will each be sitting next to a female. Therefore, f would be 2. For m, since there are four men, each is sitting next to at least one male, so m is still 6. So that gives me (2,6) again.If the two women are sitting apart, meaning there's at least one man between them, then each woman is sitting next to two men. So, each woman has two men adjacent to her. That means four different men are sitting next to at least one female. So, f would be 4. For m, since there are four men, each is sitting next to at least one male, so m is still 6. So that gives me the pair (4,6).Moving on to three women. Again, there are different configurations. If all three women are sitting together, forming a block, then the two men adjacent to the block will each be sitting next to a female. So, f would be 2. Wait, but actually, each woman in the block is sitting next to a man on one side and a woman on the other. So, the two men next to the block are sitting next to a female, but the women in the block are sitting next to other women. So, f would be 2. Hmm, but wait, each woman is sitting next to at least one female, but the question is about people sitting next to at least one female. So, the two men next to the block are sitting next to a female, so f is 2. But wait, the women themselves are also sitting next to females, but the count is about the number of people, not the number of adjacent females. So, the women are people, so if they are sitting next to at least one female, they count towards f. Wait, no, the definition is f is the number of people sitting next to at least 1 female. So, each person, regardless of gender, is counted if they have at least one female adjacent. So, in the case of three women sitting together, the two men next to the block are sitting next to a female, so they count towards f. Additionally, the women in the block are sitting next to other women, so they also count towards f because they are sitting next to at least one female. So, how many people is that? The two men next to the block and the three women in the block. So, that's 5 people. Wait, but the total number of people is 6, so one person is left. That person is a man sitting next to two men. So, f would be 5 because five people are sitting next to at least one female, and m would be 6 because everyone is sitting next to at least one male. Wait, but let me think again.Wait, if three women are sitting together, then the two men adjacent to the block are sitting next to a female, so they count towards f. The three women are sitting next to at least one female (each other), so they also count towards f. So, total f is 5. The remaining man is sitting between two men, so he doesn't count towards f, but he does count towards m because he's sitting next to males. So, m is 6 because everyone is sitting next to at least one male. So, that gives me (5,6). Hmm, but I need to check if this is correct.Alternatively, if the three women are arranged alternately with men, like woman, man, woman, man, woman, man. In this case, each woman is sitting next to two men, and each man is sitting next to two women. So, for f, every person is sitting next to at least one female because the men are next to women and the women are next to men. So, f would be 6. Similarly, m would be 6 because everyone is sitting next to at least one male. So, that gives me (6,6).Wait, so depending on how the three women are arranged, we can have either (5,6) or (6,6). Hmm, so I need to consider both cases. So, for three women, if they are all together, f is 5, and if they are alternated, f is 6.Wait, but in the case where three women are together, the two men next to the block are sitting next to a female, and the three women are sitting next to females as well. So, f is 5. The remaining man is sitting next to two men, so he doesn't contribute to f. So, f is 5, and m is 6 because everyone is sitting next to at least one male. So, that's (5,6).But wait, in the case of three women sitting together, the women are sitting next to each other, so they are next to females, but they are also people, so they count towards f because they are sitting next to at least one female. So, yes, f is 5.Okay, so for three women, we have two possibilities: (5,6) and (6,6). Hmm.Wait, but let me think again. If three women are sitting together, then the two men next to the block are sitting next to a female, so they count towards f. The three women are sitting next to at least one female (each other), so they also count towards f. So, total f is 5. The remaining man is sitting next to two men, so he doesn't count towards f. So, f is 5, and m is 6 because everyone is sitting next to at least one male. So, (5,6).If the three women are alternated with men, then every person is sitting next to at least one female and at least one male. So, f is 6 and m is 6, giving (6,6).So, for three women, we have two possible pairs: (5,6) and (6,6).Wait, but earlier, when I considered two women sitting apart, I got (4,6). So, maybe for three women, depending on the arrangement, we can have different f values.Wait, actually, when three women are sitting together, f is 5, and when they are alternated, f is 6. So, that's two different pairs.Okay, moving on to four women. Let's see. If four women are sitting together, then the two men are sitting next to the block. So, the two men are sitting next to a female, so they count towards f. The four women are sitting next to at least one female, so they also count towards f. So, f would be 6 because all six people are sitting next to at least one female. Wait, but wait, the two men are sitting next to the block of women, so they are next to a female, and the women are next to each other. So, yes, f is 6. For m, since there are two men, each is sitting next to at least one male (themselves and each other). Wait, no, the men are sitting next to women on one side and a man on the other. So, each man is sitting next to at least one male (the other man). So, m is 6 because everyone is sitting next to at least one male. So, that gives me (6,6).Alternatively, if the four women are split into two groups of two, separated by men. So, for example, two women, a man, two women, a man. In this case, each man is sitting next to two women, so they count towards f. The women are sitting next to each other and next to men. So, f would be 6 because everyone is sitting next to at least one female. Similarly, m would be 6 because everyone is sitting next to at least one male. So, that also gives me (6,6).Wait, but what if the four women are arranged with one man between them? Like woman, man, woman, man, woman, man, woman. Wait, but that's four women and two men, so it's similar to the alternated case. But in this case, each man is sitting between two women, so they are sitting next to females. The women are sitting next to men and women. So, f is 6 and m is 6.Wait, but if four women are sitting together, then the two men are sitting next to the block, so they are sitting next to a female, and the women are sitting next to each other. So, f is 6 because all six people are sitting next to at least one female. Similarly, m is 6 because everyone is sitting next to at least one male. So, regardless of how the four women are arranged, f is 6 and m is 6. So, only (6,6) for four women.Wait, but earlier, when I considered two women sitting apart, I got (4,6). So, maybe for four women, depending on the arrangement, we can have different f values. Wait, no, if four women are sitting together, f is 6. If they are split into two groups of two, separated by men, then f is still 6 because everyone is sitting next to at least one female. So, regardless of the arrangement, f is 6 for four women. So, only (6,6).Wait, but let me think again. If four women are sitting together, then the two men are sitting next to the block, so they are next to a female, and the women are next to each other. So, f is 6 because all six people are sitting next to at least one female. Similarly, m is 6 because everyone is sitting next to at least one male. So, yes, (6,6).If four women are split into two groups of two, separated by men, then each man is sitting next to two women, so they count towards f. The women are sitting next to each other and next to men. So, f is 6 because everyone is sitting next to at least one female. Similarly, m is 6 because everyone is sitting next to at least one male. So, yes, (6,6).Okay, so for four women, only (6,6).Now, five women and one man. The man is sitting between two women. So, the man is sitting next to two women, so he counts towards f. The five women are sitting next to at least one female (each other) and next to the man. So, f is 6 because everyone is sitting next to at least one female. For m, the man is sitting next to two women, but he is a male, so he counts towards m. The women are sitting next to at least one male (the man). So, m is 6 because everyone is sitting next to at least one male. So, that gives me (6,6).Finally, all six women, which we already considered, gives (6,0).Wait, so let me summarize the possible pairs I've found so far:- All men: (0,6)- One woman: (2,6)- Two women together: (2,6)- Two women apart: (4,6)- Three women together: (5,6)- Three women alternated: (6,6)- Four women: (6,6)- Five women: (6,6)- All women: (6,0)So, the unique pairs are:(0,6), (2,6), (4,6), (5,6), (6,6), (6,0)Wait, that's six unique pairs. But let me check if I missed any.Wait, when I considered three women, I had two possibilities: (5,6) and (6,6). So, that adds two more pairs.But earlier, when I considered two women, I had (2,6) and (4,6). So, that's two pairs.So, in total, the unique pairs are:(0,6), (2,6), (4,6), (5,6), (6,6), (6,0)That's six pairs.Wait, but let me think again. Is (5,6) actually possible? Because when three women are sitting together, f is 5. But is that correct?Let me visualize: three women sitting together, so they form a block. The two men adjacent to the block are sitting next to a female, so they count towards f. The three women are sitting next to each other, so they are sitting next to females, so they count towards f. The remaining man is sitting next to two men, so he doesn't count towards f. So, f is 5. And m is 6 because everyone is sitting next to at least one male. So, yes, (5,6) is a valid pair.Similarly, when three women are alternated with men, f is 6 because everyone is sitting next to at least one female, and m is 6 because everyone is sitting next to at least one male.So, yes, (6,6) is also valid.So, in total, the unique pairs are six: (0,6), (2,6), (4,6), (5,6), (6,6), (6,0).Wait, but let me check if there are any other possibilities. For example, when four women are sitting together, f is 6, and m is 6. When four women are split into two groups of two, f is still 6, and m is 6. So, no new pairs there.Similarly, for five women, f is 6, and m is 6.So, I think that's all the possibilities.Therefore, the number of possible ordered pairs (f, m) is 6.

💡Alright, so I have this problem about an ellipse, and it's divided into two parts. Let me start with part (I). First, the equation of the ellipse is given as (frac{x^2}{4} + y^2 = 1). I remember that the standard form of an ellipse is (frac{x^2}{a^2} + frac{y^2}{b^2} = 1), where (a) is the semi-major axis and (b) is the semi-minor axis. In this case, (a^2 = 4) so (a = 2), and (b^2 = 1) so (b = 1). Now, the foci of an ellipse are located at ((pm c, 0)) where (c = sqrt{a^2 - b^2}). Plugging in the values, (c = sqrt{4 - 1} = sqrt{3}). So, the foci (F_1) and (F_2) are at ((- sqrt{3}, 0)) and ((sqrt{3}, 0)) respectively.The problem states that point (P) is on the ellipse in the first quadrant, so both its (x) and (y) coordinates are positive. We're given that the dot product of vectors (overrightarrow{PF_1}) and (overrightarrow{PF_2}) is (-frac{5}{4}). Let me denote the coordinates of (P) as ((x, y)). Then, the vectors (overrightarrow{PF_1}) and (overrightarrow{PF_2}) can be written as:[overrightarrow{PF_1} = (x - (-sqrt{3}), y - 0) = (x + sqrt{3}, y)][overrightarrow{PF_2} = (x - sqrt{3}, y - 0) = (x - sqrt{3}, y)]The dot product of these two vectors is:[overrightarrow{PF_1} cdot overrightarrow{PF_2} = (x + sqrt{3})(x - sqrt{3}) + y cdot y = x^2 - 3 + y^2]We're told this equals (-frac{5}{4}), so:[x^2 + y^2 - 3 = -frac{5}{4}][x^2 + y^2 = 3 - frac{5}{4} = frac{12}{4} - frac{5}{4} = frac{7}{4}]So, we have the equation (x^2 + y^2 = frac{7}{4}). But we also know that (P) lies on the ellipse, so it must satisfy (frac{x^2}{4} + y^2 = 1). Now, I have a system of two equations:1. (x^2 + y^2 = frac{7}{4})2. (frac{x^2}{4} + y^2 = 1)Let me subtract the second equation from the first to eliminate (y^2):[x^2 + y^2 - left(frac{x^2}{4} + y^2right) = frac{7}{4} - 1][x^2 - frac{x^2}{4} = frac{3}{4}][frac{3x^2}{4} = frac{3}{4}][3x^2 = 3][x^2 = 1][x = pm 1]Since (P) is in the first quadrant, (x = 1). Now, plug this back into the ellipse equation to find (y):[frac{1^2}{4} + y^2 = 1][frac{1}{4} + y^2 = 1][y^2 = frac{3}{4}][y = pm frac{sqrt{3}}{2}]Again, since (P) is in the first quadrant, (y = frac{sqrt{3}}{2}). So, the coordinates of (P) are ((1, frac{sqrt{3}}{2})).Moving on to part (II). We have a line (l) passing through the fixed point (M(0, 2)) and intersecting the ellipse at two points (A) and (B). The angle (angle AOB) is acute, where (O) is the origin. We need to find the range of the slope (k) of line (l).First, let me consider the equation of line (l). Since it passes through (M(0, 2)), its equation can be written as (y = kx + 2), where (k) is the slope.This line intersects the ellipse (frac{x^2}{4} + y^2 = 1). Let's substitute (y = kx + 2) into the ellipse equation:[frac{x^2}{4} + (kx + 2)^2 = 1][frac{x^2}{4} + k^2x^2 + 4kx + 4 = 1][left(frac{1}{4} + k^2right)x^2 + 4kx + 3 = 0]Multiply through by 4 to eliminate the fraction:[(1 + 4k^2)x^2 + 16kx + 12 = 0]This is a quadratic in (x). For the line to intersect the ellipse at two distinct points, the discriminant must be positive. The discriminant (D) is:[D = (16k)^2 - 4 cdot (1 + 4k^2) cdot 12][D = 256k^2 - 48(1 + 4k^2)][D = 256k^2 - 48 - 192k^2][D = 64k^2 - 48]For two distinct real roots, (D > 0):[64k^2 - 48 > 0][64k^2 > 48][k^2 > frac{48}{64} = frac{3}{4}][k > frac{sqrt{3}}{2} text{ or } k < -frac{sqrt{3}}{2}]So, the slope (k) must satisfy (|k| > frac{sqrt{3}}{2}).Now, the angle (angle AOB) is acute. For an angle at the origin between points (A) and (B) to be acute, the dot product of vectors (overrightarrow{OA}) and (overrightarrow{OB}) must be positive. Let me denote points (A(x_1, y_1)) and (B(x_2, y_2)). Then:[overrightarrow{OA} cdot overrightarrow{OB} = x_1x_2 + y_1y_2 > 0]From the quadratic equation, the sum and product of roots (x_1) and (x_2) are:[x_1 + x_2 = -frac{16k}{1 + 4k^2}][x_1x_2 = frac{12}{1 + 4k^2}]Also, since (y = kx + 2), we have:[y_1 = kx_1 + 2][y_2 = kx_2 + 2]So, the product (y_1y_2) is:[(kx_1 + 2)(kx_2 + 2) = k^2x_1x_2 + 2k(x_1 + x_2) + 4]Therefore, the dot product (x_1x_2 + y_1y_2) becomes:[x_1x_2 + k^2x_1x_2 + 2k(x_1 + x_2) + 4][= (1 + k^2)x_1x_2 + 2k(x_1 + x_2) + 4]Substituting the expressions for (x_1 + x_2) and (x_1x_2):[= (1 + k^2)left(frac{12}{1 + 4k^2}right) + 2kleft(-frac{16k}{1 + 4k^2}right) + 4][= frac{12(1 + k^2)}{1 + 4k^2} - frac{32k^2}{1 + 4k^2} + 4][= frac{12 + 12k^2 - 32k^2}{1 + 4k^2} + 4][= frac{12 - 20k^2}{1 + 4k^2} + 4][= frac{12 - 20k^2 + 4(1 + 4k^2)}{1 + 4k^2}][= frac{12 - 20k^2 + 4 + 16k^2}{1 + 4k^2}][= frac{16 - 4k^2}{1 + 4k^2}]We need this to be greater than 0:[frac{16 - 4k^2}{1 + 4k^2} > 0]Since the denominator (1 + 4k^2) is always positive, the inequality simplifies to:[16 - 4k^2 > 0][16 > 4k^2][4 > k^2][k^2 < 4][|k| < 2]So, combining this with the earlier condition (|k| > frac{sqrt{3}}{2}), we get:[frac{sqrt{3}}{2} < |k| < 2]Therefore, the range of the slope (k) is:[k in left(-2, -frac{sqrt{3}}{2}right) cup left(frac{sqrt{3}}{2}, 2right)]I think that's it. Let me just recap to make sure I didn't make any mistakes. For part (I), I found the coordinates of (P) by solving the system of equations derived from the dot product condition and the ellipse equation. For part (II), I set up the line equation, substituted it into the ellipse, found the discriminant condition for two intersection points, and then used the dot product condition for the angle being acute. It all seems to check out.**Final Answer**(I) The coordinates of point (P) are (boxed{Pleft(1, frac{sqrt{3}}{2}right)}).(II) The range of the slope (k) is (boxed{left(-2, -frac{sqrt{3}}{2}right) cup left(frac{sqrt{3}}{2}, 2right)}).

💡Okay, so I have this problem where I need to distribute 7 students into two dormitories, A and B. Each dormitory has to have at least 2 students. Hmm, let me think about how to approach this.First, I guess I need to figure out all the possible ways to split 7 students into two groups, making sure neither group has fewer than 2 students. Since there are two dormitories, the groups will be assigned to A and B. Let me list out the possible distributions. The total number of students is 7, so the possible splits are:- 2 students in dorm A and 5 in dorm B- 3 students in dorm A and 4 in dorm B- 4 students in dorm A and 3 in dorm B- 5 students in dorm A and 2 in dorm BWait, but actually, since the dormitories are distinct (A and B), the splits 2-5 and 5-2 are different, right? Similarly, 3-4 and 4-3 are different distributions. So, I need to consider both possibilities for each split.Now, for each of these splits, I need to calculate the number of ways to choose the students for each dorm. This sounds like a combination problem because the order in which we select the students doesn't matter.Starting with the first split: 2 students in dorm A and 5 in dorm B. The number of ways to choose 2 students out of 7 is given by the combination formula C(n, k) = n! / (k!(n - k)!). So, C(7, 2) = 21 ways.Next, for 3 students in dorm A and 4 in dorm B, the number of ways is C(7, 3) = 35 ways.Similarly, for 4 students in dorm A and 3 in dorm B, it's also C(7, 4). Wait, but C(7, 4) is the same as C(7, 3) because of the symmetry in combinations. So, that's also 35 ways.And for 5 students in dorm A and 2 in dorm B, it's C(7, 5) which is the same as C(7, 2), so 21 ways.Now, if I add up all these possibilities: 21 (for 2-5 split) + 35 (for 3-4 split) + 35 (for 4-3 split) + 21 (for 5-2 split) = 112 ways.But wait, is there another way to think about this? Maybe using the multiplication principle. Since each student has two choices, dorm A or dorm B, the total number of ways without any restrictions would be 2^7 = 128. But this includes distributions where one dorm might have fewer than 2 students, which we don't want.So, to adjust for that, I need to subtract the cases where dorm A has fewer than 2 students or dorm B has fewer than 2 students. Let's see:- If dorm A has 0 students, then all 7 are in dorm B. There's only 1 way for this.- If dorm A has 1 student, then there are C(7, 1) = 7 ways.Similarly, for dorm B:- If dorm B has 0 students, all 7 in dorm A: 1 way.- If dorm B has 1 student, C(7, 1) = 7 ways.But wait, if I subtract all these, I might be double-counting the case where both dorms have only 1 student, but that's impossible because we have 7 students. So, actually, the cases where dorm A has 0 or 1 student are distinct from dorm B having 0 or 1 student.So, total invalid distributions are 1 (all in A) + 7 (1 in A, 6 in B) + 1 (all in B) + 7 (1 in B, 6 in A) = 16 invalid distributions.Therefore, the valid distributions would be total distributions minus invalid ones: 128 - 16 = 112.Hmm, that matches the previous method. So, both approaches give me 112 ways.But wait, let me double-check. Is there another way to think about this? Maybe using permutations or something else? I don't think so because the order within the dorms doesn't matter, only the groups themselves.Alternatively, I could think of it as assigning each student to A or B, ensuring that neither dorm is left with fewer than 2 students. So, each student has 2 choices, but we have to exclude the assignments where A has less than 2 or B has less than 2.As calculated earlier, that gives 128 - 16 = 112.Yeah, I think that's solid. Both methods lead to the same answer, so I'm confident it's 112.