answer：Okay, so I have this problem where I need to fill a 3x3 grid with numbers 1 through 9, each number used exactly once. The conditions are that in each row, the numbers increase from left to right, and in each column, the numbers decrease from top to bottom. I need to figure out how many different ways I can do this. The options given are 12, 24, 42, or 48. Hmm, let's try to break this down step by step.First, let me visualize the grid. It's a 3x3 grid, so there are 9 cells. Each row has to be increasing from left to right, which means that in the first row, the smallest number will be on the left, and the largest on the right. Similarly, for the second and third rows. On the other hand, each column has to be decreasing from top to bottom, so the largest number in each column will be at the top, and the smallest at the bottom.So, if I think about the grid, the top-left corner must be the largest number in its row and column. Wait, actually, since the first row is increasing, the first cell is the smallest in its row, but since the first column is decreasing, the first cell is the largest in its column. So, the top-left cell is the smallest in its row but the largest in its column. That seems a bit conflicting, but maybe it's possible.Wait, let me clarify. In the first row, numbers increase from left to right, so the first cell is the smallest in the first row. In the first column, numbers decrease from top to bottom, so the first cell is the largest in the first column. So, the top-left cell is both the smallest in its row and the largest in its column. That means that the number in the top-left cell must be larger than all the numbers below it in the first column but smaller than all the numbers to the right in the first row.Similarly, the top-right cell is the largest in its row (since the row is increasing) and the smallest in its column (since the column is decreasing). So, the top-right cell is the largest in the first row but the smallest in the third column.This seems a bit tricky, but maybe I can approach this by considering the possible numbers that can be placed in each cell based on these constraints.Let me think about the center cell. Since the grid is 3x3, the center cell is part of both the middle row and the middle column. The middle row has to be increasing, so the center cell is larger than the cell to its left and smaller than the cell to its right. Similarly, the middle column has to be decreasing, so the center cell is smaller than the cell above it and larger than the cell below it.So, the center cell is kind of a pivot point. It has to be larger than the cell above it and the cell to its left, and smaller than the cell below it and the cell to its right.Wait, is that correct? Let me verify. In the middle column, numbers decrease from top to bottom, so the cell above the center is larger than the center, and the cell below is smaller. In the middle row, numbers increase from left to right, so the cell to the left is smaller than the center, and the cell to the right is larger. So, yes, the center cell is sandwiched between larger numbers above and to the right, and smaller numbers below and to the left.So, the center cell must be a number that is neither too big nor too small. Maybe it's somewhere in the middle of the range 1-9. Let's think about possible numbers for the center.Since the grid has numbers 1 through 9, the center cell can't be 1 or 9 because it has to be larger than some numbers and smaller than others. Similarly, it can't be 2 or 8 because it needs to have enough numbers both above and below it in the column and to the left and right in the row.Wait, actually, let's think more carefully. The center cell has to be larger than the cell above it and the cell to its left, and smaller than the cell below it and the cell to its right. So, the cell above it must be larger than the center, and the cell below must be smaller. Similarly, the cell to the left must be smaller, and the cell to the right must be larger.Therefore, the center cell must be greater than two numbers (above and left) and less than two numbers (below and right). So, in terms of the entire grid, the center cell is the fourth or fifth number? Wait, maybe not exactly, but it's somewhere in the middle.Alternatively, perhaps I can model this as a Young tableau. A Young tableau is a matrix where numbers increase across each row and down each column. Wait, but in this case, the rows increase and the columns decrease, so it's a bit different. Maybe it's a reverse Young tableau?Wait, actually, in a standard Young tableau, rows increase to the right and columns increase downward. Here, rows increase to the right, but columns decrease downward. So, it's like a reverse in the columns. Maybe this is a kind of dual or something.But maybe I can still use some concepts from Young tableaux. The number of standard Young tableaux of a given shape is given by the hook-length formula. Maybe I can use that here.Wait, but in our case, the grid is 3x3, and the conditions are that rows increase and columns decrease. So, it's a 3x3 grid with row-increasing and column-decreasing. So, is there a formula for the number of such tableaux?Alternatively, maybe I can think of it as a permutation matrix with certain constraints.Wait, another approach: Since each row is increasing and each column is decreasing, the grid is a permutation matrix where each row is increasing and each column is decreasing. So, maybe this is equivalent to a permutation matrix that is both row-increasing and column-decreasing.Wait, but actually, the grid is filled with numbers 1 through 9, each exactly once, so it's a Latin square? No, not necessarily, because in a Latin square, each number appears exactly once in each row and column, but here, the numbers are just 1 through 9, each once, but the constraints are on the order.Wait, perhaps I can model this as a permutation matrix where the rows are increasing and columns are decreasing.Wait, maybe I can think of the grid as a matrix where each row is increasing and each column is decreasing, so it's a kind of order matrix.Alternatively, perhaps I can use the concept of a permutation matrix with certain properties.Wait, another idea: Since each row is increasing, the numbers in each row must be a combination of three numbers in increasing order. Similarly, each column must be decreasing, so each column is a combination of three numbers in decreasing order.But since all numbers from 1 to 9 are used exactly once, the entire grid is a 3x3 matrix with distinct numbers 1-9, rows increasing left to right, and columns decreasing top to bottom.This seems similar to a standard Young tableau, except that in a standard Young tableau, both rows and columns are increasing. Here, rows are increasing, columns are decreasing. So, maybe it's a kind of dual.Wait, is there a formula for the number of such matrices? I'm not sure, but maybe I can compute it by considering the possible placements of numbers.Alternatively, perhaps I can think about the grid as a permutation matrix where the rows are increasing and columns are decreasing, so each row is a set of three numbers in increasing order, and each column is a set of three numbers in decreasing order.Wait, maybe I can think of the grid as a 3x3 matrix where each row is increasing and each column is decreasing. So, the first row must be the three smallest numbers, but arranged in increasing order. Wait, no, because the columns are decreasing, so the first column must be the three largest numbers in decreasing order.Wait, that might not necessarily be the case. Let me think.Wait, the first column has to be decreasing, so the top of the first column is the largest number in that column, and the bottom is the smallest. Similarly, the first row has to be increasing, so the leftmost number is the smallest in that row, and the rightmost is the largest.So, the top-left corner is the smallest in its row and the largest in its column. So, it's the smallest in the first row and the largest in the first column.Similarly, the top-right corner is the largest in its row and the smallest in its column.So, the top-left is the smallest in its row, so it's 1? Wait, no, because it's the largest in its column, so it has to be larger than the numbers below it in the first column.Wait, so the top-left number is larger than the numbers below it in the first column, but it's also the smallest in its row. So, the top-left number is the smallest in the first row but larger than the numbers in the first column below it.So, the top-left number must be greater than the numbers in the first column below it, but less than the numbers in the first row to its right.Similarly, the top-right number is the largest in the first row, but the smallest in the third column.So, the top-right number is larger than the numbers to its left in the first row, but smaller than the numbers below it in the third column.So, perhaps the top-left number is 5, and the top-right is 5? Wait, no, because all numbers are distinct.Wait, maybe I can consider possible numbers for the top-left and top-right.Wait, the top-left number has to be larger than the numbers below it in the first column, which are two numbers. So, the top-left number must be greater than two numbers, so it can't be 1, 2, or 3, because if it's 1, it can't be greater than any number, which is impossible. If it's 2, it can only be greater than 1, but there are two numbers below it, so 2 can't be greater than both. Similarly, 3 can only be greater than 1 and 2, but if the numbers below it are, say, 4 and 5, 3 can't be greater than them. So, the top-left number must be at least 4.Similarly, the top-right number is the largest in the first row, so it's greater than the numbers to its left. It's also the smallest in the third column, so it's smaller than the numbers below it in the third column. So, the top-right number must be less than two numbers below it in the third column. So, it can't be 9, 8, or 7, because if it's 9, it can't be smaller than any number below it. If it's 8, it can only be smaller than 9, but there are two numbers below it, so 8 can't be smaller than both. Similarly, 7 can only be smaller than 8 and 9, but if the numbers below it are, say, 6 and 5, 7 can't be smaller than them. So, the top-right number must be at most 6.So, the top-left number is at least 4, and the top-right number is at most 6.So, possible top-left numbers: 4, 5, 6, 7, 8, 9.Possible top-right numbers: 1, 2, 3, 4, 5, 6.But wait, since the top-left is in the first row, which is increasing, the top-left must be less than the top-middle and top-right. Similarly, the top-right is greater than the top-middle and top-left.But also, the top-left is greater than the numbers below it in the first column, which are the middle-left and bottom-left.Similarly, the top-right is less than the numbers below it in the third column, which are the middle-right and bottom-right.So, maybe I can try to figure out the possible numbers for the top-left and top-right.Let me consider the top-left number. It must be greater than the middle-left and bottom-left numbers. So, if the top-left is 4, then the middle-left and bottom-left must be less than 4, i.e., 1, 2, or 3.Similarly, if the top-left is 5, then the middle-left and bottom-left must be less than 5, i.e., 1, 2, 3, or 4.Similarly, if the top-left is 6, then the middle-left and bottom-left must be less than 6, i.e., 1, 2, 3, 4, or 5.But also, the first row must be increasing, so the top-left is less than top-middle, which is less than top-right.Similarly, the first column is decreasing, so top-left is greater than middle-left, which is greater than bottom-left.So, let's try to think about possible assignments.Case 1: Top-left is 4.Then, the first column must have two numbers below 4, so middle-left and bottom-left must be 1, 2, or 3.Also, the first row must be increasing, so top-middle and top-right must be greater than 4.Since the top-right is at most 6, as we saw earlier, so top-right can be 5 or 6.Wait, but if top-left is 4, and top-right is 5 or 6, then the top-middle must be between 4 and 5 or 4 and 6.But let's see:If top-left is 4, top-right is 5, then top-middle must be 5? But 5 is already used as top-right. Wait, no, top-middle can be 5, but then top-right would have to be greater than 5, but we said top-right is at most 6. So, top-right could be 6.Wait, maybe it's better to think in terms of possible assignments.Alternatively, maybe I can model this as a permutation problem with constraints.Wait, another idea: Since each row is increasing and each column is decreasing, the grid is a permutation matrix where rows are increasing and columns are decreasing. So, each row is a set of three numbers in increasing order, and each column is a set of three numbers in decreasing order.This is similar to a 3x3 grid where each row is a combination of three numbers in increasing order, and each column is a combination of three numbers in decreasing order, with all numbers from 1 to 9 used exactly once.This seems like a problem that can be approached using the concept of permutation matrices with specific constraints.Wait, maybe I can think of it as a 3x3 grid where each row is a combination of three numbers in increasing order, and each column is a combination of three numbers in decreasing order.So, the problem reduces to counting the number of such grids.Wait, I recall that the number of 3x3 permutation matrices where rows are increasing and columns are decreasing is equal to the number of standard Young tableaux of shape 3x3, but with columns decreasing instead of increasing. Hmm, but standard Young tableaux have both rows and columns increasing.Wait, maybe it's related to the number of semistandard Young tableaux, but I'm not sure.Alternatively, perhaps I can use the hook-length formula for standard Young tableaux, but adjust it for the column constraints.Wait, the hook-length formula gives the number of standard Young tableaux for a given shape. For a 3x3 grid, the number is 42. Wait, that's one of the answer choices. Hmm, but in our case, the columns are decreasing, not increasing. So, is it the same as the number of standard Young tableaux?Wait, no, because in a standard Young tableau, both rows and columns are increasing. Here, rows are increasing and columns are decreasing, so it's a different structure.Wait, but maybe the number is the same because it's just a matter of orientation. If I reverse the order of the columns, perhaps it becomes a standard Young tableau. So, maybe the number is the same as the number of standard Young tableaux of shape 3x3, which is 42. So, the answer might be 42.But wait, let me verify that. The number of standard Young tableaux of a 3x3 grid is indeed 42, calculated by the hook-length formula. The hook length for each cell is as follows:For the first row, the hook lengths are 5, 4, 3.For the second row, the hook lengths are 3, 2, 1.For the third row, the hook lengths are 3, 2, 1.Wait, no, actually, the hook length for each cell is the number of cells to the right and below, including itself.So, for a 3x3 grid:First row: 5, 4, 3Second row: 3, 2, 1Third row: 3, 2, 1Wait, no, that doesn't seem right. Let me recalculate.Wait, the hook length for a cell (i,j) is the number of cells directly to the right in the same row, directly below in the same column, plus one for the cell itself.So, for a 3x3 grid:First row, first cell: hook length is 3 (right) + 3 (below) + 1 = 7? Wait, no, that's not correct.Wait, actually, for a Young diagram, the hook length is the number of cells to the right in the same row, plus the number of cells below in the same column, plus one for the cell itself.So, for the first cell (1,1):Number of cells to the right: 2Number of cells below: 2So, hook length = 2 + 2 + 1 = 5.Similarly, for cell (1,2):Number of cells to the right: 1Number of cells below: 2Hook length = 1 + 2 + 1 = 4.Cell (1,3):Number of cells to the right: 0Number of cells below: 2Hook length = 0 + 2 + 1 = 3.Second row, first cell (2,1):Number of cells to the right: 2Number of cells below: 1Hook length = 2 + 1 + 1 = 4.Wait, no, that doesn't seem right. Wait, in the second row, first column, the number of cells to the right is 2 (since it's a 3x3 grid), but the number of cells below is 1 (only the third row). So, hook length = 2 + 1 + 1 = 4.Similarly, cell (2,2):Number of cells to the right: 1Number of cells below: 1Hook length = 1 + 1 + 1 = 3.Cell (2,3):Number of cells to the right: 0Number of cells below: 1Hook length = 0 + 1 + 1 = 2.Third row, first cell (3,1):Number of cells to the right: 2Number of cells below: 0Hook length = 2 + 0 + 1 = 3.Cell (3,2):Number of cells to the right: 1Number of cells below: 0Hook length = 1 + 0 + 1 = 2.Cell (3,3):Number of cells to the right: 0Number of cells below: 0Hook length = 0 + 0 + 1 = 1.So, the hook lengths for the 3x3 grid are:5 4 34 3 23 2 1Then, the number of standard Young tableaux is given by the hook-length formula:Number of SYTs = 9! / (5*4*3*4*3*2*3*2*1)Let me compute that.First, compute the denominator:5 * 4 * 3 * 4 * 3 * 2 * 3 * 2 * 1Let me compute step by step:5 * 4 = 2020 * 3 = 6060 * 4 = 240240 * 3 = 720720 * 2 = 14401440 * 3 = 43204320 * 2 = 86408640 * 1 = 8640So, denominator is 8640.Numerator is 9! = 362880.So, 362880 / 8640 = let's compute that.Divide numerator and denominator by 10: 36288 / 864Divide numerator and denominator by 8: 4536 / 108Divide numerator and denominator by 12: 378 / 9 = 42.So, the number of standard Young tableaux is 42. So, that's one of the answer choices, option C.But wait, in our problem, the columns are decreasing instead of increasing. So, is the number of such tableaux the same as the number of standard Young tableaux?Wait, in a standard Young tableau, both rows and columns are increasing. In our case, rows are increasing, columns are decreasing. So, it's a different structure.But is there a bijection between such tableaux and standard Young tableaux? Maybe by reversing the order of the columns or something.Wait, if I reverse the order of the columns, then the columns that were decreasing become increasing. So, if I have a grid where rows are increasing and columns are decreasing, and I reverse the order of the columns, I get a grid where rows are increasing (since reversing columns doesn't affect row order) and columns are increasing. So, that would be a standard Young tableau.Therefore, the number of such grids is equal to the number of standard Young tableaux of shape 3x3, which is 42. So, the answer is 42, which is option C.Wait, but let me think again. Is there a bijection between the two? If I reverse the columns, does that map each grid with rows increasing and columns decreasing to a standard Young tableau?Yes, because reversing the columns would turn the decreasing columns into increasing columns, while the rows remain increasing because reversing columns doesn't affect the row order.Therefore, the number of such grids is equal to the number of standard Young tableaux, which is 42.So, I think the answer is 42, which is option C.**Final Answer**boxed{C}

answer：Okay, so I have this equation: (3x - m)/2 - (x + m)/3 = 5/6. There are two parts to the problem. Let me tackle them one by one.Starting with part (1): If m = -1, find the solution to the equation.Alright, so m is given as -1. I need to substitute m = -1 into the equation and solve for x. Let me write down the equation with m replaced by -1.So, substituting m = -1, the equation becomes:(3x - (-1))/2 - (x + (-1))/3 = 5/6Simplify the numerators:(3x + 1)/2 - (x - 1)/3 = 5/6Hmm, okay. Now, to solve this equation, I think I should eliminate the denominators first. The denominators are 2, 3, and 6. The least common denominator (LCD) here is 6. So, I can multiply both sides of the equation by 6 to get rid of the fractions.Multiplying each term by 6:6*(3x + 1)/2 - 6*(x - 1)/3 = 6*(5/6)Simplify each term:6/2 is 3, so 3*(3x + 1) = 9x + 3Similarly, 6/3 is 2, so 2*(x - 1) = 2x - 2And on the right side, 6*(5/6) is just 5.So, putting it all together:9x + 3 - (2x - 2) = 5Wait, hold on. I have to remember that the second term is subtracted, so I need to distribute the negative sign.So, expanding that:9x + 3 - 2x + 2 = 5Combine like terms:9x - 2x is 7x3 + 2 is 5So, 7x + 5 = 5Now, subtract 5 from both sides:7x = 0Divide both sides by 7:x = 0Hmm, so x equals 0 when m is -1. Let me check if that makes sense by plugging x = 0 and m = -1 back into the original equation.Original equation:(3*0 - (-1))/2 - (0 + (-1))/3 = ?Simplify:(0 + 1)/2 - (0 - 1)/3 = 1/2 - (-1)/3 = 1/2 + 1/3Convert to common denominator:3/6 + 2/6 = 5/6Which is equal to the right side of the equation, 5/6. So, yes, x = 0 is correct.Alright, part (1) is done. Now, moving on to part (2): If x = 5 is a solution to the equation, find the value of (1/2)m² + 2m.So, this time, x is given as 5, and I need to find m first, then compute (1/2)m² + 2m.Let me substitute x = 5 into the original equation:(3*5 - m)/2 - (5 + m)/3 = 5/6Simplify the numerators:(15 - m)/2 - (5 + m)/3 = 5/6Again, I can eliminate the denominators by multiplying both sides by 6, the LCD.Multiplying each term by 6:6*(15 - m)/2 - 6*(5 + m)/3 = 6*(5/6)Simplify each term:6/2 is 3, so 3*(15 - m) = 45 - 3m6/3 is 2, so 2*(5 + m) = 10 + 2mRight side: 6*(5/6) is 5Putting it all together:45 - 3m - (10 + 2m) = 5Again, distribute the negative sign:45 - 3m - 10 - 2m = 5Combine like terms:45 - 10 is 35-3m - 2m is -5mSo, 35 - 5m = 5Subtract 35 from both sides:-5m = 5 - 35-5m = -30Divide both sides by -5:m = 6Okay, so m is 6. Now, I need to compute (1/2)m² + 2m.Let me plug m = 6 into that expression:(1/2)*(6)² + 2*(6)First, compute 6 squared: 6*6 = 36Then, (1/2)*36 = 18Next, 2*6 = 12Add them together: 18 + 12 = 30So, the value is 30.Let me double-check my steps to make sure I didn't make a mistake.Starting with x = 5:(15 - m)/2 - (5 + m)/3 = 5/6Multiply by 6:3*(15 - m) - 2*(5 + m) = 5Which is 45 - 3m -10 -2m = 5Simplify: 35 -5m =5Subtract 35: -5m = -30Divide: m=6Then, compute (1/2)*36 +12=18+12=30. Yep, that seems correct.So, both parts are done. I think I got them right.**Final Answer**(1) The solution is boxed{0}.(2) The value is boxed{30}.

answer：Okay, so I have this probability problem here about selecting flight attendants. Let me try to understand it step by step. The problem says: For a transatlantic flight, three flight attendants are selected by lot from 20 girls competing for these positions. Seven of them are blondes, and the rest are brunettes. What is the probability that among the three chosen flight attendants there will be at least one blonde and at least one brunette?Alright, so we have 20 girls total. Out of these, 7 are blondes, which means the remaining 13 must be brunettes. We need to find the probability that when we select 3 girls, there is at least one blonde and at least one brunette. Hmm, okay. So, probability problems can sometimes be tricky, but I remember that when dealing with "at least one" scenarios, it's often helpful to use the complementary probability. That is, instead of calculating the probability of the event happening directly, we calculate the probability of the event not happening and subtract it from 1.So, in this case, the event we're interested in is having at least one blonde and at least one brunette. The complementary event would be having all blondes or all brunettes. So, if I can find the probability of selecting all blondes or all brunettes, then subtract that from 1, I should get the desired probability.Let me write that down:Probability(at least one blonde and one brunette) = 1 - Probability(all blondes or all brunettes)But wait, actually, since all blondes and all brunettes are mutually exclusive events (you can't have all blondes and all brunettes at the same time), I can calculate each probability separately and then add them together.So, Probability(all blondes or all brunettes) = Probability(all blondes) + Probability(all brunettes)Therefore, the formula becomes:Probability(at least one blonde and one brunette) = 1 - [Probability(all blondes) + Probability(all brunettes)]Alright, now I need to calculate Probability(all blondes) and Probability(all brunettes).Since the selection is done by lot, I assume it's a random selection without replacement. So, this is a hypergeometric probability problem, but since we're dealing with combinations, I can use combinations to calculate the probabilities.First, let's figure out the total number of ways to choose 3 girls out of 20. That's the denominator for all our probabilities.Total combinations = C(20, 3)Where C(n, k) is the combination formula, which is n! / (k! * (n - k)! )Calculating that:C(20, 3) = 20! / (3! * 17!) = (20 * 19 * 18) / (3 * 2 * 1) = 1140So, there are 1140 possible ways to choose 3 girls from 20.Now, let's calculate the number of ways to choose all blondes. There are 7 blondes, so we need to choose all 3 from them.Number of ways to choose all blondes = C(7, 3)Calculating that:C(7, 3) = 7! / (3! * 4!) = (7 * 6 * 5) / (3 * 2 * 1) = 35Similarly, the number of ways to choose all brunettes. There are 13 brunettes, so we need to choose all 3 from them.Number of ways to choose all brunettes = C(13, 3)Calculating that:C(13, 3) = 13! / (3! * 10!) = (13 * 12 * 11) / (3 * 2 * 1) = 286So, the number of ways to choose all blondes is 35, and all brunettes is 286. Therefore, the total number of ways to choose either all blondes or all brunettes is 35 + 286 = 321.Therefore, the probability of selecting all blondes or all brunettes is 321 / 1140.Let me compute that:321 divided by 1140. Let me see, both numbers are divisible by 3.321 ÷ 3 = 1071140 ÷ 3 = 380So, 107 / 380. Hmm, can this be simplified further? Let's check.107 is a prime number, I believe, because it doesn't divide by 2, 3, 5, 7, 11. Let's see: 107 ÷ 13 is about 8.23, which isn't an integer. So, 107 is prime. 380 is 38 * 10, which is 2 * 19 * 2 * 5. So, no common factors. So, 107/380 is the simplified fraction.So, the probability of all blondes or all brunettes is 107/380.Therefore, the probability of having at least one blonde and one brunette is 1 - 107/380.Calculating that:1 is equal to 380/380, so 380/380 - 107/380 = (380 - 107)/380 = 273/380.Now, let's see if 273 and 380 have any common factors.273: Let's factor it. 273 ÷ 3 = 91. So, 3 * 91. 91 is 7 * 13. So, 273 = 3 * 7 * 13.380: As above, 380 = 2 * 2 * 5 * 19.So, no common factors between 273 and 380. Therefore, 273/380 is the simplified fraction.So, the probability is 273/380. Let me check if that can be simplified further or if I made a mistake in calculations.Wait, 273 divided by 3 is 91, and 380 divided by 3 is not an integer. So, no, it can't be simplified further.Alternatively, as a decimal, 273 divided by 380 is approximately 0.7184, so about 71.84%.Wait, let me double-check my calculations because 35 + 286 is 321, which is correct. 321 divided by 1140 is 0.2816, so 28.16%. Therefore, 1 - 0.2816 is 0.7184, which is about 71.84%. That seems reasonable.Alternatively, I can think of it as: the number of favorable outcomes is total combinations minus all blondes minus all brunettes, which is 1140 - 35 - 286 = 1140 - 321 = 819.Wait, hold on, 1140 - 321 is 819? Wait, 1140 - 300 is 840, then subtract 21 more is 819. So, 819 favorable outcomes.Therefore, probability is 819 / 1140.Simplify that: 819 ÷ 3 = 273, 1140 ÷ 3 = 380. So, 273/380, which is the same as before.So, that's consistent.Alternatively, maybe I can compute it using probabilities step by step.Probability of at least one blonde and at least one brunette is equal to 1 - Probability(all blondes) - Probability(all brunettes).We already calculated Probability(all blondes) as C(7,3)/C(20,3) = 35/1140 ≈ 0.0307.Probability(all brunettes) is C(13,3)/C(20,3) = 286/1140 ≈ 0.2509.Adding those together: 0.0307 + 0.2509 ≈ 0.2816, which is approximately 28.16%.Subtracting from 1: 1 - 0.2816 ≈ 0.7184, which is 71.84%.So, that's consistent with the previous calculation.Alternatively, maybe I can compute it using another method, like considering the different cases.So, the favorable cases are when we have 1 blonde and 2 brunettes, or 2 blondes and 1 brunette.So, the number of ways to choose 1 blonde and 2 brunettes is C(7,1)*C(13,2).Similarly, the number of ways to choose 2 blondes and 1 brunette is C(7,2)*C(13,1).Therefore, total favorable outcomes = C(7,1)*C(13,2) + C(7,2)*C(13,1).Let me compute each term.First, C(7,1) is 7, and C(13,2) is (13*12)/2 = 78. So, 7 * 78 = 546.Second, C(7,2) is (7*6)/2 = 21, and C(13,1) is 13. So, 21 * 13 = 273.Therefore, total favorable outcomes = 546 + 273 = 819.Which is the same as before. So, 819 / 1140 = 273 / 380 ≈ 0.7184.So, that's consistent.Therefore, the probability is 273/380, which is approximately 71.84%.Wait, just to make sure, let me verify the calculations once more.Total number of ways: C(20,3) = 1140. Correct.Number of all blondes: C(7,3) = 35. Correct.Number of all brunettes: C(13,3) = 286. Correct.Total unfavorable: 35 + 286 = 321. Correct.Favorable: 1140 - 321 = 819. Correct.Probability: 819 / 1140 = 273 / 380. Correct.Yes, that seems solid.Alternatively, thinking in terms of probability multiplication:The probability that the first person is a blonde is 7/20. Then, the second is also blonde: 6/19, and the third is blonde: 5/18. So, the probability of all blondes is (7/20)*(6/19)*(5/18) = (7*6*5)/(20*19*18) = 210 / 6840 = 7 / 228 ≈ 0.0307.Similarly, probability of all brunettes: first is brunette: 13/20, second: 12/19, third: 11/18. So, (13/20)*(12/19)*(11/18) = (13*12*11)/(20*19*18) = 1716 / 6840 = 1716 ÷ 12 = 143, 6840 ÷12=570. So, 143/570 ≈ 0.2509.Adding these together: 7/228 + 143/570. Let me compute that.First, find a common denominator. 228 and 570.228 factors: 2*2*3*19.570 factors: 2*3*5*19.So, the least common multiple is 2*3*5*19*2 = 1140.So, convert both fractions to denominator 1140.7/228 = (7*5)/1140 = 35/1140.143/570 = (143*2)/1140 = 286/1140.So, total is 35 + 286 = 321 / 1140, which is the same as before.So, 321/1140 = 107/380 ≈ 0.2816.Therefore, 1 - 0.2816 = 0.7184, which is 273/380.So, same result.Therefore, I can be confident that the probability is 273/380.Alternatively, if I wanted to write this as a decimal, it's approximately 0.7184, which is about 71.84%.So, to sum up, the probability that among the three chosen flight attendants there will be at least one blonde and at least one brunette is 273/380, or approximately 71.84%.I think that's thorough enough. I considered multiple methods: combinations, complementary probability, step-by-step probability multiplication, and verified each step. All approaches led me to the same answer, so I feel confident that this is correct.**Final Answer**The probability is boxed{dfrac{273}{380}}.

answer：To determine the smallest possible perimeter of a scalene triangle with integer side lengths, I need to identify three distinct integers that satisfy the triangle inequality theorem. This theorem states that the sum of the lengths of any two sides must be greater than the length of the third side.I'll start by considering the smallest possible integers for the sides. Let's begin with the smallest integer, 1, but quickly realize that a triangle with sides 1, 2, and 3 doesn't satisfy the triangle inequality because 1 + 2 is not greater than 3.Next, I'll try the set of sides 2, 3, and 4. Checking the triangle inequality:- 2 + 3 > 4 (5 > 4)- 2 + 4 > 3 (6 > 3)- 3 + 4 > 2 (7 > 2)All conditions are satisfied, so a triangle with sides 2, 3, and 4 is valid. Adding these together, the perimeter is 2 + 3 + 4 = 9.To ensure this is the smallest possible perimeter, I'll check if there's a smaller set of integers that meets the criteria. The next smallest set would be 1, 2, and 3, but as previously determined, it doesn't satisfy the triangle inequality. Therefore, the smallest possible perimeter for a scalene triangle with integer side lengths is 9.

answer：Okay, so I have this problem here: The standard deviation of the numbers a₁, a₂, a₃, a₄, a₅ is 2. I need to find the variance of the numbers 3a₁ - 2, 3a₂ - 2, 3a₃ - 2, 3a₄ - 2, 3a₅ - 2. Hmm, let me think about how to approach this.First, I remember that standard deviation is a measure of how spread out the numbers are, and variance is just the square of the standard deviation. So if the standard deviation is 2, then the variance must be 2 squared, which is 4. That seems straightforward.But now, the question is about transforming each of the original numbers by multiplying them by 3 and then subtracting 2. I need to figure out how this transformation affects the variance. I think there's a property of variance that relates to linear transformations of data. Let me recall what that is.I remember that if you have a set of data and you apply a linear transformation to each data point, such as y = a*x + b, the variance of the new data set is a squared times the variance of the original data set. The constant term, b, doesn't affect the variance because it just shifts all the data points by the same amount, which doesn't change how spread out they are. So, in this case, since we're multiplying each a_i by 3 and then subtracting 2, the variance should be affected only by the multiplication by 3.So, applying that property, the variance of the transformed data should be (3)² times the original variance. Since the original variance is 4, that would be 9 * 4, which is 36. Therefore, the variance of the numbers 3a₁ - 2, 3a₂ - 2, etc., should be 36.Wait, let me make sure I'm not making a mistake here. Sometimes, I get confused between how transformations affect mean and variance. The mean would be affected by both the multiplication and the subtraction, right? Because if you have y = a*x + b, the new mean is a times the original mean plus b. But variance is only affected by the scaling factor, not the shift. So, subtracting 2 doesn't change the spread, it just moves everything down by 2 units. So, yeah, only the multiplication by 3 affects the variance.Let me also think about why that is. Variance measures the average squared deviation from the mean. So, if you multiply each data point by 3, each deviation from the mean is also multiplied by 3. Then, when you square that deviation, it becomes 9 times the original squared deviation. Hence, the variance increases by a factor of 9. Subtracting 2 doesn't change the deviations from the mean because it subtracts the same amount from each data point, so the deviations remain the same. Therefore, the variance is only scaled by the square of the multiplication factor.Just to double-check, let's consider an example. Suppose we have a simple data set: 1, 2, 3. The mean is 2, and the variance is [(1-2)² + (2-2)² + (3-2)²]/3 = (1 + 0 + 1)/3 = 2/3. Now, if we apply the transformation y = 3x - 2, the new data set is 1, 4, 7. The new mean is (1 + 4 + 7)/3 = 12/3 = 4. The deviations from the mean are (1-4) = -3, (4-4)=0, (7-4)=3. Squaring these gives 9, 0, 9. The variance is (9 + 0 + 9)/3 = 18/3 = 6. The original variance was 2/3, and 3 squared is 9, so 9*(2/3) is indeed 6. That checks out.So, applying this to the problem at hand, since the original variance is 4, multiplying each term by 3 increases the variance by 9, giving 36. Subtracting 2 doesn't change it. Therefore, the variance of the transformed numbers is 36.I think that makes sense. I don't see any mistakes in my reasoning. The key points are understanding how linear transformations affect variance and remembering that only the scaling factor affects variance, while the constant shift doesn't. It's also helpful to test it with a simple example to confirm.**Final Answer**The variance of the transformed numbers is boxed{36}.

answer：Okay, so I have this problem where a stepwise right triangle was drawn on graph paper with legs equal to 6 cells each. Then, all the grid lines inside the triangle were outlined. I need to find the maximum number of rectangles that can be found in this drawing. Hmm, okay. Let me try to visualize this.First, a right triangle with legs of 6 cells each. So, if I imagine graph paper, the triangle would have its right angle at the origin, and the legs would extend 6 units along the x-axis and y-axis. Since it's stepwise, I think that means it's made up of small squares, each one cell, connected in a step-like manner. So, each step would be a square, and the triangle would be built up from these squares.Wait, actually, if it's a right triangle with legs of 6 cells, then the hypotenuse would be the line connecting (6,0) to (0,6). But since it's stepwise, maybe it's constructed by connecting points along the grid lines, creating a series of small right triangles or squares? Hmm, I might need to clarify that.But perhaps it's more straightforward. Maybe it's a right triangle where each leg is divided into 6 equal segments, each of one cell length. So, the triangle is divided into smaller right triangles, each with legs of 1 cell. But then, the grid lines inside the triangle are outlined. So, we have a grid of lines inside the triangle, forming a sort of triangular lattice.Wait, but the question is about rectangles. So, in this grid, how many rectangles can we find? Hmm. Normally, in a rectangular grid, the number of rectangles is calculated by choosing two horizontal lines and two vertical lines, which form a rectangle. But in a triangular grid, it's a bit different because the lines aren't all horizontal and vertical; they might be at 45 degrees or something.Wait, hold on. Maybe the triangle is actually a square grid that's been cut along the diagonal, making a right triangle. So, if the original square grid is 6x6, then the triangle would have 6 cells along each leg, and the grid lines inside would form a sort of stepped pattern.Yes, that makes sense. So, the triangle is essentially half of a 6x6 grid, with the hypotenuse being the diagonal from (0,6) to (6,0). So, inside this triangle, the grid lines are outlined, meaning both the horizontal, vertical, and diagonal lines are drawn? Or just the horizontal and vertical?Wait, the problem says "all grid lines inside the triangle were outlined." So, in the original square grid, the grid lines are horizontal and vertical. When we take the triangle, we still have horizontal and vertical grid lines inside it, but the diagonal is the hypotenuse. So, the grid lines inside the triangle are just the horizontal and vertical lines that lie within the triangle.So, in that case, the figure is a right triangle with legs of 6 units, divided into 1x1 cells, with grid lines only horizontal and vertical, not the diagonals. So, to find the number of rectangles, we need to consider the horizontal and vertical lines inside this triangle.But wait, in a triangle, the number of horizontal and vertical lines is not uniform. For example, at the base, which is the x-axis from (0,0) to (6,0), there are 6 cells. Then, as we go up each unit, the number of cells decreases by one. So, at y=1, the horizontal line would go from (0,1) to (5,1), which is 5 cells. Similarly, at y=2, it's from (0,2) to (4,2), which is 4 cells, and so on until y=6, which is just a single cell at (0,6).Similarly, the vertical lines would go from (x,0) up to (x,6 - x), since the hypotenuse is at x + y = 6. So, for each x from 0 to 6, the vertical line at x goes up to y = 6 - x.So, in this setup, how do we count the number of rectangles? Normally, in a full grid, the number of rectangles is calculated by choosing two vertical lines and two horizontal lines. The number of rectangles is then the combination of vertical lines taken two at a time multiplied by the combination of horizontal lines taken two at a time.But in this case, since it's a triangle, the number of vertical and horizontal lines varies depending on the position. So, we can't just use the standard formula. Instead, we need to find a way to count all possible rectangles within this triangular grid.Let me think. A rectangle is defined by its top, bottom, left, and right sides. So, in this grid, each rectangle must have its top and bottom sides as horizontal grid lines and its left and right sides as vertical grid lines.But in the triangle, the horizontal lines are of different lengths. So, for a rectangle to exist, the horizontal lines it uses must span from some x1 to x2, and the vertical lines must span from some y1 to y2, such that the intersection points (x1,y1), (x2,y1), (x1,y2), (x2,y2) all lie within the triangle.Wait, but in the triangle, the horizontal lines at each y level have different lengths. So, for each y, the horizontal line goes from x=0 to x=6 - y. Similarly, each vertical line at x goes from y=0 to y=6 - x.So, to form a rectangle, we need two horizontal lines and two vertical lines such that their intersection points are all within the triangle.Alternatively, maybe we can model this as a grid where each row has a certain number of cells, decreasing as we go up, and each column has a certain number of cells, decreasing as we go to the right.So, perhaps we can think of this as a kind of stepped grid, where each row has one less cell than the row below it, and each column has one less cell than the column to its left.In such a grid, the number of rectangles can be calculated by considering all possible combinations of two horizontal lines and two vertical lines that form a rectangle. But because the grid is not uniform, we have to adjust our counting accordingly.Let me try to approach this step by step.First, let's consider the horizontal lines. There are 7 horizontal lines (including the base) at y=0, y=1, ..., y=6. Similarly, there are 7 vertical lines at x=0, x=1, ..., x=6.But each horizontal line at y=k has a length of (6 - k) cells. So, the number of vertical units (i.e., the number of vertical lines) that intersect with this horizontal line is (6 - k + 1). Similarly, each vertical line at x=k has a length of (6 - k) cells, so the number of horizontal units (i.e., the number of horizontal lines) that intersect with this vertical line is (6 - k + 1).Wait, maybe that's complicating things. Let's think differently.In the full grid, the number of rectangles is given by (n+1 choose 2) * (m+1 choose 2), where n and m are the number of rows and columns. But in our case, it's a triangular grid, so it's not a full grid.Alternatively, perhaps we can model this as a grid where each row has a decreasing number of cells. So, row 0 (the base) has 6 cells, row 1 has 5 cells, row 2 has 4 cells, and so on up to row 6, which has 1 cell.Similarly, each column x has (6 - x) cells. So, column 0 has 6 cells, column 1 has 5 cells, ..., column 6 has 1 cell.In such a grid, the number of rectangles can be calculated by considering all possible pairs of rows and columns that can form the top, bottom, left, and right sides of a rectangle.Wait, but in a non-uniform grid, the number of rectangles isn't straightforward. Maybe we can use the concept of counting rectangles in a grid with varying row and column lengths.I recall that in a grid where each row has a certain number of cells, the number of rectangles can be calculated by summing over all possible pairs of rows and columns.Specifically, for each pair of horizontal lines (top and bottom), we can determine the number of vertical lines that span between them, and then compute the number of rectangles between those two horizontal lines.Similarly, for each pair of vertical lines (left and right), we can determine the number of horizontal lines that span between them, and compute the number of rectangles between those two vertical lines.But I think it's more efficient to approach this by considering each possible rectangle's height and width.Wait, another approach: The number of rectangles in a grid is equal to the sum over all possible heights and widths of the number of positions where such a rectangle can fit.In our case, since the grid is triangular, the number of positions for a rectangle of a certain height and width depends on where it is placed.Alternatively, perhaps we can model this as a grid where each cell is identified by its row and column, but the grid is only defined for cells where row + column <= 6.Wait, that might be a useful way to think about it. So, each cell is at position (i, j) where i is the row number (starting from 0 at the top) and j is the column number (starting from 0 at the left). But in our case, the triangle is such that for each row i, the number of cells is (6 - i). So, the cells are where i + j <= 6.Wait, actually, if we consider the triangle with legs along the x and y axes, then each cell is at position (x, y) where x and y are integers such that x + y <= 6.So, the grid is a set of points (x, y) where x and y are integers from 0 to 6, and x + y <= 6.In this case, the number of rectangles is the number of axis-aligned rectangles that can be formed within this set of points.An axis-aligned rectangle is defined by its bottom-left corner (x1, y1) and top-right corner (x2, y2), where x2 > x1 and y2 > y1, and all four corners are within the set.So, to count the number of rectangles, we need to count all possible pairs of (x1, y1) and (x2, y2) such that x1 < x2, y1 < y2, and x2 + y2 <= 6.Wait, but actually, since the grid is a right triangle, the condition is that x2 + y2 <= 6, but also x1 + y1 <= 6, and x2 + y1 <= 6, and x1 + y2 <= 6.Wait, no, actually, the rectangle must lie entirely within the triangle. So, all four corners must satisfy x + y <= 6.Therefore, for a rectangle with bottom-left corner (x1, y1) and top-right corner (x2, y2), we must have:x1 < x2,y1 < y2,x2 + y2 <= 6,x1 + y2 <= 6,x2 + y1 <= 6,x1 + y1 <= 6.But since x1 < x2 and y1 < y2, the last condition is automatically satisfied if the others are.So, the key conditions are:1. x2 + y2 <= 6,2. x1 + y2 <= 6,3. x2 + y1 <= 6.So, given that, how can we count the number of such rectangles?Perhaps we can iterate over all possible x1, x2, y1, y2 and count the number of valid rectangles.But that might be too time-consuming. Maybe we can find a formula or a combinatorial approach.Alternatively, let's consider that in order to form a rectangle, we need two distinct x-coordinates and two distinct y-coordinates such that the rectangle they form is entirely within the triangle.So, the number of rectangles is equal to the number of ways to choose two x-coordinates and two y-coordinates such that the rectangle defined by them is inside the triangle.So, let's denote the x-coordinates as x1 and x2, with x1 < x2, and y-coordinates as y1 and y2, with y1 < y2.We need to ensure that:x2 + y2 <= 6,x1 + y2 <= 6,x2 + y1 <= 6.So, for given x1 and x2, what is the range of y1 and y2 such that these conditions are satisfied?Alternatively, for given y1 and y2, what is the range of x1 and x2?This seems complicated, but maybe we can fix x1 and x2 first and then find the possible y1 and y2.Let me try that.Suppose we fix x1 and x2, where 0 <= x1 < x2 <=6.Then, for these x1 and x2, we need to find the number of pairs y1 and y2 such that:y1 < y2,y2 <= 6 - x2,y2 <= 6 - x1,y1 <= 6 - x2,y1 <= 6 - x1.Wait, let's see:From condition x2 + y2 <=6, we get y2 <=6 -x2.From condition x1 + y2 <=6, we get y2 <=6 -x1.Similarly, from x2 + y1 <=6, we get y1 <=6 -x2.From x1 + y1 <=6, we get y1 <=6 -x1.But since x1 < x2, 6 -x1 >6 -x2.So, the stricter condition is y2 <=6 -x2 and y1 <=6 -x2.But y1 must be less than y2, so y1 can range from 0 up to y2 -1, but y2 is at most 6 -x2.Wait, perhaps it's better to think of it as for given x1 and x2, the maximum y2 is min(6 -x1, 6 -x2). But since x1 <x2, 6 -x1 >6 -x2, so min is 6 -x2. So, y2 can go up to 6 -x2.Similarly, y1 can go up to 6 -x2 as well, but y1 must be less than y2.So, for given x1 and x2, the number of possible y1 and y2 is the number of pairs where y1 < y2 <=6 -x2.Which is equal to C(6 -x2 +1, 2). Because y2 can be from 1 to 6 -x2, and y1 can be from 0 to y2 -1.Wait, actually, the number of pairs (y1, y2) with y1 < y2 <= N is C(N +1, 2). Because y2 can be from 1 to N, and for each y2, y1 can be from 0 to y2 -1, which is y2 choices. So, total is sum_{y2=1}^N y2 = N(N+1)/2 = C(N+1, 2).But in our case, N is 6 -x2.So, the number of y pairs is C(6 -x2 +1, 2) = C(7 -x2, 2).Therefore, for each pair x1 <x2, the number of rectangles is C(7 -x2, 2).But wait, is that correct? Let me check.Suppose x2 is fixed, say x2=3. Then, 6 -x2=3, so y2 can be up to 3. Then, the number of y pairs is C(4,2)=6.Which would correspond to y2=1, y1=0; y2=2, y1=0,1; y2=3, y1=0,1,2. So, total 1+2+3=6, which is correct.So, yes, for each x2, the number of y pairs is C(7 -x2, 2).But wait, does x1 affect this? Because we fixed x1 and x2, but in the above reasoning, x1 only affects the upper bound on y2 through 6 -x1, but since 6 -x1 >6 -x2, the limiting factor is 6 -x2.Therefore, for each x2, regardless of x1, as long as x1 <x2, the number of y pairs is C(7 -x2, 2).But wait, that can't be, because x1 also affects the maximum y1 and y2.Wait, no, actually, for a given x1 and x2, the maximum y2 is 6 -x2, as 6 -x1 is larger, so it doesn't impose a stricter condition. Similarly, y1 can go up to 6 -x2 as well.Therefore, for each x1 <x2, the number of y pairs is C(7 -x2, 2).But wait, that would mean that for each x2, the number of rectangles contributed by all x1 <x2 is (x2) * C(7 -x2, 2). Because x1 can be from 0 to x2 -1, which is x2 choices.Wait, no, actually, for each x2, the number of x1 is x2 (since x1 can be 0,1,...,x2 -1). For each such x1, the number of y pairs is C(7 -x2, 2). So, total rectangles for a given x2 is x2 * C(7 -x2, 2).Therefore, the total number of rectangles is the sum over x2 from 1 to 6 of [x2 * C(7 -x2, 2)].Wait, let's test this with x2=1.x2=1: x1 can be 0. Number of y pairs is C(6, 2)=15. So, total rectangles: 1*15=15.x2=2: x1 can be 0,1. Number of y pairs is C(5,2)=10. So, total rectangles: 2*10=20.x2=3: x1=0,1,2. y pairs=C(4,2)=6. Total:3*6=18.x2=4: x1=0,1,2,3. y pairs=C(3,2)=3. Total:4*3=12.x2=5: x1=0,1,2,3,4. y pairs=C(2,2)=1. Total:5*1=5.x2=6: x1=0,1,2,3,4,5. y pairs=C(1,2)=0, since C(1,2)=0. So, total:6*0=0.So, total rectangles would be 15+20+18+12+5+0=70.Wait, so the total number of rectangles is 70?But let me think again. Is this correct?Wait, when x2=6, y pairs would be C(1,2)=0, which makes sense because y2 can only be up to 0, which is not possible since y2 must be at least y1 +1. So, no rectangles for x2=6.Similarly, for x2=5, y2 can be up to 1, so y pairs are C(2,2)=1, which is correct: y1=0, y2=1.So, the calculation seems correct.Therefore, the total number of rectangles is 15+20+18+12+5=70.Wait, but let me cross-verify this with another approach.Another way to think about this is that in the triangular grid, each rectangle is determined by choosing two distinct x-coordinates and two distinct y-coordinates such that the rectangle lies entirely within the triangle.So, the number of rectangles is equal to the number of such quadruples (x1, x2, y1, y2) with x1 <x2, y1 <y2, and x2 + y2 <=6.So, we can model this as counting the number of 2x2 subgrids within the triangular grid.Alternatively, perhaps we can use combinatorics.Let me consider that for each possible width w and height h, the number of rectangles of size w x h is equal to the number of positions where such a rectangle can fit.In our case, the width w is the difference x2 -x1, and the height h is y2 -y1.But since the grid is triangular, the number of positions for a rectangle of size w x h depends on w and h.Wait, perhaps for a rectangle of width w and height h, the number of positions is equal to the number of x1 such that x1 + w <=6 - y1, and similarly for y1.But this seems complicated.Alternatively, maybe we can use the formula for the number of rectangles in a grid where each row has a certain number of cells.I found a resource that says the number of rectangles in a grid with m rows and n columns is m*n*(m+1)*(n+1)/4. But that's for a full grid.Wait, no, actually, the standard formula is (m choose 2)*(n choose 2). So, for m rows and n columns, it's C(m,2)*C(n,2).But in our case, it's a triangular grid, so it's not a full grid.Wait, perhaps we can model this as a grid where each row has a decreasing number of cells, and each column has a decreasing number of cells.I found a similar problem online where the number of rectangles in a right triangular grid is calculated as the sum from i=1 to n of sum from j=1 to n of (n - i +1)(n - j +1) - but I'm not sure.Wait, no, that might not be correct.Alternatively, perhaps we can think of the grid as a combination of smaller grids.Wait, another approach: The number of rectangles in the triangle is equal to the number of rectangles in the full 6x6 grid minus the number of rectangles that lie entirely above the diagonal.But that might not be straightforward because the diagonal cuts through the grid, so some rectangles would be partially above and partially below.Alternatively, perhaps we can use the fact that the number of rectangles in the triangle is equal to the sum from k=1 to 6 of the number of rectangles in each row.But each row has a different number of cells.Wait, actually, in the triangle, each row y has (6 - y) cells. So, the number of rectangles that have their base on row y is C(6 - y, 2). But that's only for rectangles that are 1 unit high.Wait, no, actually, for each row y, the number of rectangles that span from row y to row y + h is C(6 - y - h +1, 2).Wait, this is getting too convoluted.Wait, let's go back to the initial approach where we fixed x2 and calculated the number of rectangles for each x2. We got a total of 70 rectangles.But let me check with a smaller case to see if this approach is correct.Suppose the triangle has legs of 2 cells each. So, it's a smaller triangle.Using the same method:For x2=1: x1=0. y pairs=C(6 -1,2)=C(5,2)=10? Wait, no, wait, in the smaller case, 6 would be 2.Wait, actually, in the original problem, the legs are 6 cells, so in the smaller case, legs are 2 cells.So, for legs=2:x2 can be 1 and 2.For x2=1: x1=0. y pairs=C(2 -1 +1,2)=C(2,2)=1.So, total rectangles:1*1=1.For x2=2: x1=0,1. y pairs=C(2 -2 +1,2)=C(1,2)=0.So, total rectangles:2*0=0.Total rectangles=1+0=1.But in reality, in a 2x2 triangle, how many rectangles are there?Let me visualize it.The triangle has 3 horizontal lines: y=0 (2 cells), y=1 (1 cell), y=2 (0 cells).Similarly, vertical lines: x=0 (2 cells), x=1 (1 cell), x=2 (0 cells).So, the grid points are:(0,0), (1,0), (2,0)(0,1), (1,1)(0,2)So, the rectangles are:- The unit square at (0,0) to (1,1). So, only 1 rectangle.Which matches our calculation of 1.So, the method works for n=2.Another test case: n=3.Using the same method:x2 can be 1,2,3.For x2=1: x1=0. y pairs=C(3 -1 +1,2)=C(3,2)=3.So, total rectangles:1*3=3.For x2=2: x1=0,1. y pairs=C(3 -2 +1,2)=C(2,2)=1.Total rectangles:2*1=2.For x2=3: x1=0,1,2. y pairs=C(3 -3 +1,2)=C(1,2)=0.Total rectangles:3*0=0.Total rectangles=3+2+0=5.Now, let's count manually.In a 3x3 triangle, the grid points are:(0,0), (1,0), (2,0), (3,0)(0,1), (1,1), (2,1)(0,2), (1,2)(0,3)So, the rectangles are:1x1: There are several.At the base:- (0,0)-(1,1)- (1,0)-(2,1)- (2,0)-(3,1)But wait, (3,0) can't form a rectangle because there's no cell above it.Wait, actually, in the triangle, the top row is y=3, which is just (0,3). So, the rectangles are:- (0,0)-(1,1)- (1,0)-(2,1)- (2,0)-(3,1)- (0,1)-(1,2)- (1,1)-(2,2)So, total 5 rectangles, which matches our calculation.Therefore, the method seems to be correct.So, returning to our original problem with n=6, the total number of rectangles is 70.Wait, but let me check with another approach.Another way to think about it is that the number of rectangles in such a triangular grid is equal to the sum from k=1 to n of k*(k+1)*(2k+1)/8, but I'm not sure.Wait, actually, I found a formula online that the number of rectangles in a right triangular grid of size n is n(n+1)(n+2)(3n+1)/24.But let me test this formula with n=2.n=2: 2*3*4*7/24= 168/24=7. But we know that for n=2, the number of rectangles is 1. So, this formula is incorrect.Wait, maybe it's another formula.Alternatively, perhaps the number of rectangles is equal to the sum from i=1 to n of sum from j=1 to n of (n - i +1)(n - j +1) - but that seems too large.Wait, no, in our initial approach, for n=6, we got 70 rectangles.Wait, let me check with n=3, which gave us 5 rectangles.Using the formula from the initial approach:Sum over x2=1 to 3 of [x2 * C(4 -x2, 2)].Wait, for n=3, 6 -x2 would be 3 -x2 +1?Wait, no, in the original problem, it was 6 -x2, but for n=3, it should be 3 -x2.Wait, maybe I confused the variables.Wait, in the original problem, the legs are 6 cells, so the formula was x2 * C(7 -x2, 2). For n=3, it would be x2 * C(4 -x2, 2).So, for n=3:x2=1: 1*C(3,2)=3x2=2:2*C(2,2)=2x2=3:3*C(1,2)=0Total=3+2+0=5, which matches.Similarly, for n=2:x2=1:1*C(2,2)=1x2=2:2*C(1,2)=0Total=1+0=1, which matches.So, the formula seems to hold.Therefore, for n=6, the total number of rectangles is 70.Wait, but let me compute it step by step again to be sure.For x2=1:x1=0. y pairs=C(6,2)=15.Rectangles:1*15=15.x2=2:x1=0,1. y pairs=C(5,2)=10.Rectangles:2*10=20.x2=3:x1=0,1,2. y pairs=C(4,2)=6.Rectangles:3*6=18.x2=4:x1=0,1,2,3. y pairs=C(3,2)=3.Rectangles:4*3=12.x2=5:x1=0,1,2,3,4. y pairs=C(2,2)=1.Rectangles:5*1=5.x2=6:x1=0,1,2,3,4,5. y pairs=C(1,2)=0.Rectangles:6*0=0.Total:15+20=35; 35+18=53; 53+12=65; 65+5=70; 70+0=70.Yes, that's consistent.Therefore, the maximum number of rectangles that can be found in this drawing is 70.**Final Answer**The maximum number of rectangles is boxed{70}.