answer：Okay, so I have this problem here: In an acute-angled triangle ABC, the area is 10√3. The lengths of sides AB and AC are 5 and 8 respectively. I need to find the length of BC. Hmm, let's see how to approach this.First, I remember that the area of a triangle can be calculated using the formula: (1/2)*base*height. But in this case, I don't know the height corresponding to any of the sides. Alternatively, I also recall that the area can be found using the formula involving two sides and the sine of the included angle: (1/2)*ab*sinθ, where a and b are two sides, and θ is the angle between them.Since I know sides AB and AC, which are 5 and 8, maybe I can use this formula. Let me denote angle BAC as θ. Then, the area would be (1/2)*5*8*sinθ = 20*sinθ. According to the problem, the area is 10√3. So, I can set up the equation:20*sinθ = 10√3Dividing both sides by 20:sinθ = (10√3)/20 = √3/2Okay, so sinθ = √3/2. That means θ is either 60 degrees or 120 degrees because sine of both these angles is √3/2. But wait, the triangle is acute-angled, which means all angles are less than 90 degrees. So, θ must be 60 degrees.Alright, so angle BAC is 60 degrees. Now, I need to find the length of BC. Since I know two sides and the included angle, I can use the Law of Cosines. The Law of Cosines states that for any triangle with sides a, b, c opposite angles A, B, C respectively, c² = a² + b² - 2ab*cos(C).In this case, BC is the side opposite angle A, which is 60 degrees. So, BC² = AB² + AC² - 2*AB*AC*cosθ.Plugging in the values:BC² = 5² + 8² - 2*5*8*cos60°Calculating each term:5² = 258² = 642*5*8 = 80cos60° = 0.5So, BC² = 25 + 64 - 80*0.5Compute 80*0.5: that's 40.So, BC² = 25 + 64 - 40Adding 25 and 64: 89Then subtract 40: 89 - 40 = 49So, BC² = 49Taking the square root of both sides: BC = 7Wait, that seems straightforward. Let me double-check my steps to make sure I didn't make a mistake.1. Calculated the area using (1/2)*ab*sinθ: correct.2. Set up the equation 20*sinθ = 10√3: correct.3. Solved for sinθ: √3/2, which gives θ = 60° or 120°, but since the triangle is acute, θ must be 60°: correct.4. Applied the Law of Cosines: BC² = 5² + 8² - 2*5*8*cos60°: correct.5. Calculated each term: 25, 64, 80, cos60°=0.5: correct.6. Subtracted 40 from 89: got 49, so BC=7: correct.Hmm, seems solid. But just to be thorough, let me consider if there's another way to approach this problem, maybe using Heron's formula or coordinate geometry, to see if I get the same answer.Let's try Heron's formula. Heron's formula states that the area of a triangle is √[s(s-a)(s-b)(s-c)], where s is the semi-perimeter, and a, b, c are the sides.But wait, I don't know all three sides, so I can't directly apply Heron's formula. Maybe I can set up an equation with Heron's formula and solve for BC. Let me denote BC as x.So, sides are AB=5, AC=8, BC=x. The semi-perimeter s = (5 + 8 + x)/2 = (13 + x)/2.Then, the area squared is [s(s - a)(s - b)(s - c)] = [ (13 + x)/2 * ( (13 + x)/2 - 5 ) * ( (13 + x)/2 - 8 ) * ( (13 + x)/2 - x ) ]Simplify each term:s = (13 + x)/2s - a = (13 + x)/2 - 5 = (13 + x - 10)/2 = (3 + x)/2s - b = (13 + x)/2 - 8 = (13 + x - 16)/2 = (x - 3)/2s - c = (13 + x)/2 - x = (13 + x - 2x)/2 = (13 - x)/2So, the area squared is:[ (13 + x)/2 * (3 + x)/2 * (x - 3)/2 * (13 - x)/2 ]Which simplifies to:[(13 + x)(13 - x)(x + 3)(x - 3)] / 16Notice that (13 + x)(13 - x) = 169 - x² and (x + 3)(x - 3) = x² - 9.So, the numerator becomes (169 - x²)(x² - 9).Therefore, area squared is [(169 - x²)(x² - 9)] / 16.But the area is given as 10√3, so area squared is (10√3)² = 100*3 = 300.So, set up the equation:[(169 - x²)(x² - 9)] / 16 = 300Multiply both sides by 16:(169 - x²)(x² - 9) = 4800Let me expand the left side:(169 - x²)(x² - 9) = 169x² - 1521 - x⁴ + 9x² = (169x² + 9x²) - x⁴ - 1521 = 178x² - x⁴ - 1521So, 178x² - x⁴ - 1521 = 4800Bring all terms to one side:-x⁴ + 178x² - 1521 - 4800 = 0Simplify:-x⁴ + 178x² - 6321 = 0Multiply both sides by -1 to make it positive:x⁴ - 178x² + 6321 = 0Let me let y = x², so the equation becomes:y² - 178y + 6321 = 0Now, solve for y using quadratic formula:y = [178 ± √(178² - 4*1*6321)] / 2Compute discriminant:178² = 316844*1*6321 = 25284So, discriminant is 31684 - 25284 = 6400√6400 = 80Thus, y = [178 ± 80]/2Compute both possibilities:First, y = (178 + 80)/2 = 258/2 = 129Second, y = (178 - 80)/2 = 98/2 = 49So, y = 129 or y = 49Since y = x², so x² = 129 or x² = 49Thus, x = √129 or x = 7But x is the length BC, so it has to be positive. So, possible solutions are x=7 or x=√129≈11.357.But wait, in the triangle, the sum of two sides must be greater than the third side. So, let's check both possibilities.First, if BC=7:Check triangle inequalities:AB + BC > AC: 5 + 7 = 12 > 8: yes.AC + BC > AB: 8 + 7 = 15 > 5: yes.AB + AC > BC: 5 + 8 = 13 > 7: yes.So, BC=7 is valid.Now, if BC=√129≈11.357:Check triangle inequalities:AB + BC > AC: 5 + 11.357 ≈16.357 >8: yes.AC + BC > AB: 8 +11.357≈19.357>5: yes.AB + AC > BC: 5 +8=13 >11.357: yes.So, both solutions satisfy the triangle inequality.But wait, the triangle is acute-angled. So, we need to check which of these solutions result in an acute triangle.Hmm, how do we check if a triangle is acute? A triangle is acute if all its angles are less than 90 degrees. Alternatively, using the Law of Cosines, for all sides, the square of each side should be less than the sum of the squares of the other two sides.So, for BC=7:Check if 7² < 5² +8²: 49 <25 +64=89: yes, 49<89. So, angle opposite BC is acute.Similarly, check for other sides:5²=25 <8² +7²=64+49=113: yes.8²=64 <5² +7²=25+49=74: 64 <74: yes.So, all angles are acute. So, BC=7 is valid.Now, check BC=√129≈11.357:Check if (√129)²=129 <5² +8²=25+64=89: 129 <89? No, 129>89. So, the angle opposite BC would be obtuse.But the triangle is given as acute-angled, so BC cannot be √129. Therefore, BC must be 7.So, that confirms my earlier result.Alternatively, another way to think about this is using coordinates. Let me place point A at the origin (0,0), point B at (5,0), since AB=5. Then, point C is somewhere in the plane such that AC=8 and the area is 10√3.The area can also be calculated as (1/2)*base*height. If I take AB as the base, which is 5 units, then the height from C to AB must satisfy (1/2)*5*h =10√3, so h= (10√3)*2 /5=4√3.So, the height from point C to AB is 4√3. Since AB is along the x-axis from (0,0) to (5,0), the y-coordinate of point C must be 4√3. So, point C is at (x, 4√3) for some x.Now, AC=8, so the distance from A(0,0) to C(x,4√3) is 8.So, sqrt(x² + (4√3)²)=8Compute (4√3)²=16*3=48So, sqrt(x² +48)=8Square both sides: x² +48=64Thus, x²=16, so x=4 or x=-4.But since we're dealing with a triangle, point C can be on either side of AB. However, since the triangle is acute, we need to ensure that all angles are less than 90 degrees.If x=4, then point C is at (4,4√3). Let's compute the lengths of BC.Point B is at (5,0), point C is at (4,4√3). So, distance BC is sqrt[(5-4)² + (0 -4√3)²] = sqrt[1 + 48]=sqrt[49]=7.If x=-4, point C is at (-4,4√3). Then, distance BC is sqrt[(5 - (-4))² + (0 -4√3)²] = sqrt[9² + (4√3)²] = sqrt[81 +48]=sqrt[129]≈11.357.But as before, when BC=√129, the triangle is obtuse, so we discard this solution.Therefore, BC=7.So, through multiple methods: using the area formula with sine, Law of Cosines, Heron's formula, and coordinate geometry, I consistently get BC=7. Therefore, I'm confident that the answer is 7.**Final Answer**The length of BC is boxed{7}.

answer：Okay, so I have this probability problem with two parts. Let me try to tackle them one by one. Starting with part (1): There are three cards in a box. One card is red on both sides, another is black on both sides, and the third has one red side and one black side. I need to find the probability that the other side is also red given that a randomly shown side is red.Hmm, this seems like a conditional probability problem. I remember that conditional probability is about the probability of an event given that another event has occurred. The formula is P(A|B) = P(A and B) / P(B). Let me define the events:- Let A be the event that the card drawn is the double red card.- Let B be the event that a red side is shown.We need to find P(A|B), which is the probability that the card is the double red card given that a red side is shown.So, according to the formula, I need to find P(A and B) and P(B).First, P(A and B): If we have the double red card, then showing a red side is certain. So, P(B|A) is 1. The probability of drawing the double red card is 1/3. Therefore, P(A and B) = P(A) * P(B|A) = (1/3) * 1 = 1/3.Next, P(B): The probability that a red side is shown. There are three cards, each with two sides, so a total of 6 sides. Out of these, the double red card has 2 red sides, the double black has 0 red sides, and the mixed card has 1 red side. So, total red sides are 2 + 0 + 1 = 3. Therefore, the probability of showing a red side is 3/6 = 1/2.Wait, is that correct? Let me think again. Since each card is equally likely to be drawn, and then a side is randomly shown. So, the probability is actually the sum over each card of the probability of drawing that card times the probability of showing a red side given that card.So, P(B) = P(drawing red-red) * P(red|red-red) + P(drawing black-black) * P(red|black-black) + P(drawing red-black) * P(red|red-black)Which is (1/3)*1 + (1/3)*0 + (1/3)*(1/2) = (1/3) + 0 + (1/6) = 1/2. Yeah, same result.So, P(A|B) = (1/3) / (1/2) = 2/3.Wait, that seems high, but I think it's correct. Because when you see a red side, it's more likely that it's the double red card since it has two red sides.Let me double-check. There are three red sides: two from the double red and one from the mixed card. So, if you pick a red side, the chance it's from the double red is 2/3, and from the mixed is 1/3. So, yes, the probability that the other side is red is 2/3. That makes sense.Okay, so part (1) is 2/3.Moving on to part (2): If a card is randomly drawn, one side is shown, returned to the box, and then another card is drawn and one side is shown. The number of times black is shown is denoted as X. I need to find the distribution and expected value of X.So, X can be 0, 1, or 2. Because in two trials, each time a side is shown, which can be black or red. So, the number of blacks can be 0, 1, or 2.First, let's find the distribution, which is the probability mass function of X. So, P(X=0), P(X=1), and P(X=2).To find these probabilities, I need to model the two trials. Each trial is independent because after each draw, the card is returned to the box. So, the two trials are identical and independent.First, let's find the probability that a single trial results in a black side. Let's denote this as p.So, similar to part (1), p is the probability that a randomly shown side is black. Let's compute that.Again, using the same reasoning as part (1), but now for black sides.Total sides: 6. Number of black sides: the double black card has 2, the mixed card has 1, and the double red has 0. So, total black sides: 2 + 1 + 0 = 3. So, p = 3/6 = 1/2.Wait, same as red. So, each trial has a 1/2 chance of showing black and 1/2 chance of showing red.But wait, is that correct? Because the process is: draw a card, show a random side, return the card. So, each trial is independent, and each has the same probability distribution.So, each trial is a Bernoulli trial with success probability p = 1/2 (success being showing black). Therefore, X is the sum of two independent Bernoulli trials, so it follows a Binomial distribution with parameters n=2 and p=1/2.Therefore, the distribution of X is:- P(X=0) = C(2,0)*(1/2)^0*(1/2)^2 = 1*1*1/4 = 1/4- P(X=1) = C(2,1)*(1/2)^1*(1/2)^1 = 2*(1/2)*(1/2) = 1/2- P(X=2) = C(2,2)*(1/2)^2*(1/2)^0 = 1*(1/4)*1 = 1/4So, the distribution is:X | 0 | 1 | 2P(X) | 1/4 | 1/2 | 1/4Now, the expected value E[X] is n*p = 2*(1/2) = 1.Alternatively, since each trial has expectation 1/2, the total expectation is 1/2 + 1/2 = 1.Wait, but let me make sure that the trials are indeed independent and each has p=1/2.Is the probability of showing black in a single trial really 1/2?Let me compute it again.Total sides: 6.Number of black sides: 3 (2 from double black, 1 from mixed).So, when you draw a card and show a random side, the probability of black is 3/6 = 1/2. So yes, each trial is a Bernoulli(1/2).Therefore, X ~ Binomial(2, 1/2), so the distribution is as above, and E[X] = 1.Wait, but hold on a second. Is the process of drawing a card, showing a side, and returning the card to the box, independent each time? Yes, because each time you start fresh, so the two trials are independent.Therefore, yes, the number of blacks shown, X, is the sum of two independent Bernoulli trials, each with p=1/2, so it's Binomial(2,1/2).Therefore, the distribution is P(X=0)=1/4, P(X=1)=1/2, P(X=2)=1/4, and the expected value is 1.Wait, but let me think again. Is there a different way to compute P(X=0), P(X=1), P(X=2) without assuming the trials are Bernoulli?Maybe I should compute it explicitly, just to make sure.So, each trial is: draw a card, show a side, return the card.So, for each trial, the possible outcomes are red or black, each with probability 1/2.Therefore, two trials, each with two possible outcomes, so four possible outcomes:1. Red, Red2. Red, Black3. Black, Red4. Black, BlackEach outcome has probability (1/2)*(1/2)=1/4.Now, X is the number of blacks in the two trials.So:- X=0 corresponds to outcome 1: Red, Red. So, probability 1/4.- X=1 corresponds to outcomes 2 and 3: Red, Black and Black, Red. Each has probability 1/4, so total 1/2.- X=2 corresponds to outcome 4: Black, Black. Probability 1/4.So, same as above. So, yes, the distribution is correct.Therefore, the expected value is 0*(1/4) + 1*(1/2) + 2*(1/4) = 0 + 1/2 + 1/2 = 1.So, that's consistent.Therefore, part (2) is that X follows a Binomial distribution with parameters n=2 and p=1/2, so P(X=0)=1/4, P(X=1)=1/2, P(X=2)=1/4, and E[X]=1.Wait, but just to make sure, is there any dependency between the two trials? Since after each trial, the card is returned to the box, so the trials are independent. So, yes, the two trials are independent, so the number of blacks is indeed Binomial.Alternatively, if the card wasn't returned, it might be different, but since it is returned, it's independent.Therefore, I think my reasoning is correct.So, summarizing:(1) The probability that the other side is also red is 2/3.(2) The distribution of X is P(X=0)=1/4, P(X=1)=1/2, P(X=2)=1/4, and the expected value is 1.**Final Answer**(1) boxed{dfrac{2}{3}}(2) The distribution of (X) is (P(X=0) = dfrac{1}{4}), (P(X=1) = dfrac{1}{2}), (P(X=2) = dfrac{1}{4}), and the expected value is boxed{1}.

answer：Okay, so I have this problem where a light ray starts at point A, which is at (0, 1) on the y-axis. It reflects off a point C on the x-axis and then passes through point B at (3, 3). I need to find the length of the path from A to B via the reflection at C. Hmm, reflections... I remember something about using the method of images to solve these kinds of problems. Let me think about how that works.First, when dealing with reflections, especially in optics, the law of reflection states that the angle of incidence is equal to the angle of reflection. But sometimes, instead of dealing with angles, it's easier to use the method of images, where you reflect one of the points across the reflecting surface and then find a straight line path. That should give the shortest path that the light would take, considering the reflection.In this case, the light starts at A, reflects off the x-axis at point C, and then goes to B. So, the reflecting surface is the x-axis. I think I can reflect point B across the x-axis to get a new point, let's call it B'. Then, the path from A to C to B should be equivalent to the straight line path from A to B'. That makes sense because the reflection would make the path appear straight in the mirrored image.Let me write down the coordinates. Point A is (0, 1), and point B is (3, 3). Reflecting B across the x-axis would invert its y-coordinate. So, B' would be at (3, -3). Now, if I can find the straight line distance from A to B', that should give me the total path length from A to C to B.But wait, before I jump into calculating the distance, I should make sure that point C is indeed on the x-axis. So, the line from A to B' should intersect the x-axis at some point C, which is the reflection point. So, first, I can find the equation of the line connecting A (0, 1) and B' (3, -3). Then, find where this line intersects the x-axis (where y=0). That intersection point will be point C.Let me find the equation of the line AB'. The slope (m) of the line connecting A and B' can be calculated using the formula:m = (y2 - y1) / (x2 - x1)Plugging in the coordinates:m = (-3 - 1) / (3 - 0) = (-4)/3So, the slope is -4/3. Now, using the point-slope form of the equation of a line, which is:y - y1 = m(x - x1)Using point A (0, 1) as (x1, y1):y - 1 = (-4/3)(x - 0)Simplifying:y = (-4/3)x + 1Now, to find the point C where this line intersects the x-axis, I set y = 0:0 = (-4/3)x + 1Solving for x:(-4/3)x = -1Multiply both sides by (-3/4):x = (-1) * (-3/4) = 3/4So, point C is at (3/4, 0). That seems reasonable. Now, to find the total path length from A to C to B, I can calculate the distance from A to C and then from C to B, and add them together. Alternatively, since we used the method of images, the distance from A to B' is the same as the total path length from A to C to B.Let me verify both ways.First, calculating the distance from A to C:Point A is (0, 1), point C is (3/4, 0).Distance formula:d1 = sqrt[(x2 - x1)^2 + (y2 - y1)^2]d1 = sqrt[(3/4 - 0)^2 + (0 - 1)^2] = sqrt[(9/16) + (1)] = sqrt[(9/16) + (16/16)] = sqrt[25/16] = 5/4Then, calculating the distance from C to B:Point C is (3/4, 0), point B is (3, 3).Distance formula:d2 = sqrt[(3 - 3/4)^2 + (3 - 0)^2] = sqrt[(9/4)^2 + (3)^2] = sqrt[(81/16) + (9)] = sqrt[(81/16) + (144/16)] = sqrt[225/16] = 15/4So, total distance is d1 + d2 = 5/4 + 15/4 = 20/4 = 5.Alternatively, calculating the distance from A to B' directly:Point A is (0, 1), point B' is (3, -3).Distance formula:d = sqrt[(3 - 0)^2 + (-3 - 1)^2] = sqrt[(9) + (16)] = sqrt[25] = 5.So, both methods give the same result, which is reassuring.Therefore, the length of the path taken by the light from point A to point B via reflection at point C is 5 units.Wait, just to make sure I didn't make any calculation errors. Let me double-check the distances.From A (0,1) to C (3/4, 0):Difference in x: 3/4 - 0 = 3/4Difference in y: 0 - 1 = -1So, squared differences: (9/16) and (1). Sum is 25/16, square root is 5/4. Correct.From C (3/4, 0) to B (3, 3):Difference in x: 3 - 3/4 = 9/4Difference in y: 3 - 0 = 3Squared differences: (81/16) and (9). Sum is 81/16 + 144/16 = 225/16, square root is 15/4. Correct.Total: 5/4 + 15/4 = 20/4 = 5. Correct.And the distance from A to B' is sqrt[(3)^2 + (-4)^2] = 5. Correct.So, all steps check out. I think I'm confident that the total path length is 5 units.**Final Answer**The length of the path taken by the light from point A to point B is boxed{5}.

answer：First, I recognize that Rachel has 1 toy.Since John has 6 more toys than Rachel, I add 6 to Rachel's toys to find that John has 7 toys.Jason has three times as many toys as John. Therefore, I multiply John's number of toys by 3 to determine that Jason has 21 toys.

answer：Okay, so I have this problem about a sequence {a_n} where S_n is the sum of the first n terms. The problem says that for n ≥ 2, the point (a_{n-1}, 2a_n) lies on the line y = 2x + 1. Also, the first term a_1 is the minimum value of the quadratic function y = x² - 2x + 3. I need to find S_9, which is the sum of the first 9 terms.First, let me break down the problem step by step.1. **Finding a_1:** The first term a_1 is the minimum value of the quadratic function y = x² - 2x + 3. I remember that for a quadratic function in the form y = ax² + bx + c, the vertex (which gives the minimum if a > 0) occurs at x = -b/(2a). So, for y = x² - 2x + 3, a = 1, b = -2. Therefore, the x-coordinate of the vertex is -(-2)/(2*1) = 2/2 = 1. Plugging x = 1 back into the function to find the minimum value: y = (1)² - 2*(1) + 3 = 1 - 2 + 3 = 2. So, a_1 = 2.2. **Understanding the relationship for n ≥ 2:** The point (a_{n-1}, 2a_n) lies on the line y = 2x + 1. That means when we plug in x = a_{n-1}, y should be equal to 2a_n. So, substituting into the equation: 2a_n = 2*a_{n-1} + 1. Simplifying this equation: 2a_n = 2a_{n-1} + 1 Divide both sides by 2: a_n = a_{n-1} + 1/2 So, this tells me that each term a_n is equal to the previous term a_{n-1} plus 1/2. That means the sequence {a_n} is an arithmetic sequence with common difference d = 1/2.3. **Confirming the type of sequence:** Since a_n = a_{n-1} + 1/2 for n ≥ 2, and a_1 is given, this is indeed an arithmetic sequence. The common difference is 1/2.4. **Finding the general term a_n:** For an arithmetic sequence, the nth term is given by: a_n = a_1 + (n - 1)*d Plugging in the known values: a_n = 2 + (n - 1)*(1/2) Simplify: a_n = 2 + (n/2 - 1/2) a_n = 2 - 1/2 + n/2 a_n = 3/2 + n/2 Alternatively, I can write this as: a_n = (n + 3)/2 Let me check this formula with n = 1: a_1 = (1 + 3)/2 = 4/2 = 2. Correct. For n = 2: a_2 = (2 + 3)/2 = 5/2. Let's verify using the recursive relation: a_2 = a_1 + 1/2 = 2 + 1/2 = 5/2. Correct. For n = 3: a_3 = (3 + 3)/2 = 6/2 = 3. Using recursion: a_3 = a_2 + 1/2 = 5/2 + 1/2 = 6/2 = 3. Correct. So, the formula seems to hold.5. **Calculating S_9:** S_n is the sum of the first n terms of an arithmetic sequence. The formula for the sum is: S_n = n/2 * (a_1 + a_n) Alternatively, since we have the formula for a_n, we can use: S_n = n/2 * [2a_1 + (n - 1)d] Let me compute both ways to verify. First, using S_n = n/2 * (a_1 + a_n): We know a_1 = 2, and a_9 = (9 + 3)/2 = 12/2 = 6. So, S_9 = 9/2 * (2 + 6) = (9/2)*8 = 9*4 = 36. Alternatively, using S_n = n/2 * [2a_1 + (n - 1)d]: S_9 = 9/2 * [2*2 + (9 - 1)*(1/2)] = 9/2 * [4 + 8*(1/2)] = 9/2 * [4 + 4] = 9/2 * 8 = 9*4 = 36. Both methods give the same result, so that's reassuring.6. **Double-checking the recursive relation:** Just to make sure I didn't make a mistake earlier, let's compute a few terms manually and see if the sum adds up. a_1 = 2 a_2 = 5/2 = 2.5 a_3 = 3 a_4 = 3.5 a_5 = 4 a_6 = 4.5 a_7 = 5 a_8 = 5.5 a_9 = 6 Now, let's add these up: 2 + 2.5 = 4.5 4.5 + 3 = 7.5 7.5 + 3.5 = 11 11 + 4 = 15 15 + 4.5 = 19.5 19.5 + 5 = 24.5 24.5 + 5.5 = 30 30 + 6 = 36 Yep, adding them term by term also gives S_9 = 36. So that's consistent.7. **Alternative approach using recursion:** Another way to approach this is to recognize that since each term increases by 1/2, starting from 2, the sequence is 2, 2.5, 3, 3.5, ..., up to the 9th term. The number of terms is 9, the first term is 2, the last term is 6. The average of the first and last term is (2 + 6)/2 = 4. Multiply by the number of terms: 4*9 = 36. Same result.8. **Conclusion:** All methods lead to S_9 = 36. Therefore, I'm confident that the answer is 36.**Final Answer**The sum of the first 9 terms is boxed{36}.

answer：Okay, so I have this problem: There are ___ different ways to assign 3 teachers to teach in 6 classes, with each teacher teaching 2 distinct classes. Hmm, let me try to figure this out step by step.First, I need to understand what exactly is being asked. We have 3 teachers, and 6 classes. Each teacher needs to be assigned to 2 distinct classes. So, each teacher will be responsible for two different classes, and all 6 classes will be covered by these 3 teachers. That makes sense.So, essentially, we're looking to distribute 6 classes among 3 teachers, where each teacher gets exactly 2 classes. The question is about the number of different ways this can be done. I think this is a combinatorial problem, involving combinations and permutations.Let me break it down. Maybe I can think of it as partitioning the 6 classes into 3 groups of 2 classes each, and then assigning each group to a teacher. But wait, the teachers are distinct, right? So, the order in which we assign these groups matters.Alternatively, maybe I can think of it as assigning each class to a teacher, but each teacher must end up with exactly 2 classes. Hmm, that might be another approach.Let me consider the first approach: partitioning the 6 classes into 3 groups of 2, and then assigning each group to a teacher. So, how many ways can we partition 6 classes into 3 groups of 2?I remember that when partitioning a set into groups, the number of ways is given by the multinomial coefficient. For partitioning into groups of sizes k1, k2, ..., kn, the formula is 6! / (k1! * k2! * ... * kn!). In this case, each group has size 2, so it would be 6! / (2! * 2! * 2!). But wait, since the groups are indistinct at this point, meaning that swapping the groups doesn't create a new partition, do we need to adjust for that?Wait, actually, no. Because in this case, the groups are going to be assigned to distinct teachers, so the order of the groups matters. So, maybe we don't need to divide by the number of permutations of the groups. Let me think.If the groups were indistinct, meaning that swapping them doesn't matter, we would divide by 3! to account for the permutations of the groups. But since each group is going to be assigned to a specific teacher, who is distinct, we don't need to do that. So, the number of ways to partition the classes into groups is 6! / (2! * 2! * 2!) = 720 / (8) = 90.But wait, hold on. If we have 6 classes, and we want to divide them into 3 groups of 2, the formula is indeed 6! / (2!^3 * 3!). Wait, now I'm confused. Because sometimes when you have identical groups, you divide by the factorial of the number of groups.Let me recall the formula for partitioning into groups. If you have n items and you want to divide them into groups of sizes k1, k2, ..., km, where the groups are indistinct, the number of ways is n! / (k1! * k2! * ... * km! * m!). But in this case, since the groups are going to be assigned to distinct teachers, the groups are distinguishable. So, we don't have to divide by m!.So, in this case, the number of ways to partition the 6 classes into 3 groups of 2 is 6! / (2! * 2! * 2!) = 720 / 8 = 90. Then, since each group is assigned to a specific teacher, and the teachers are distinct, we don't need to do anything else. So, the total number of ways is 90.But wait, another thought: maybe I should consider the assignment as a permutation. Let me think differently. Suppose we assign classes to teachers one by one.First, choose 2 classes out of 6 for the first teacher. Then, choose 2 classes out of the remaining 4 for the second teacher. Then, the last 2 classes go to the third teacher. Since the teachers are distinct, the order in which we assign the classes matters.So, the number of ways would be C(6,2) * C(4,2) * C(2,2). Let's compute that:C(6,2) is 15, C(4,2) is 6, and C(2,2) is 1. So, 15 * 6 * 1 = 90. Hmm, same result as before. So, that seems consistent.Alternatively, if I think about permutations. Assigning classes to teachers can also be thought of as arranging the classes in a certain order and then grouping them. But I think the combination approach is more straightforward here.Wait, another angle: if we consider each class being assigned to a teacher, with each teacher getting exactly 2 classes. So, it's like a surjective function from the set of classes to the set of teachers, where each teacher has exactly 2 classes. The number of such functions is equal to the number of ways to partition the classes into 3 groups of 2, multiplied by the number of ways to assign these groups to the teachers.But since the teachers are distinct, the assignment is just a multiplication by 3! if the groups were indistinct. Wait, but in our earlier calculation, we didn't have to do that because we considered the groups as ordered.Wait, maybe I'm overcomplicating. Let me see.If I first partition the classes into 3 groups of 2, which can be done in 6! / (2!^3) = 720 / 8 = 90 ways. Then, since each group is assigned to a distinct teacher, and the teachers are distinct, each partition corresponds to exactly one assignment. So, the total number is 90.Alternatively, if I think of it as arranging the classes in order and assigning the first two to teacher 1, next two to teacher 2, and last two to teacher 3. But in that case, the number of arrangements would be 6! = 720, but since the order within each pair doesn't matter, we divide by 2! for each teacher, so 720 / (2! * 2! * 2!) = 90. So, same result.Therefore, I think 90 is the correct answer.But wait, let me make sure I'm not missing something. Is there a different interpretation of the problem?The problem says: assign 3 teachers to teach in 6 classes, with each teacher teaching 2 distinct classes.So, each teacher is assigned to 2 distinct classes, and all 6 classes are covered. So, yes, that's exactly what I considered.Alternatively, if the classes are distinct, and the teachers are distinct, then the number of assignments is indeed 90.Wait, another thought: is the order of classes within a teacher's assignment important? For example, does it matter which class is first or second for a teacher? The problem says "each teacher teaching 2 distinct classes," so I think the order doesn't matter. So, it's combinations, not permutations.Therefore, the calculation using combinations is correct.So, to recap, the number of ways is C(6,2) * C(4,2) * C(2,2) = 15 * 6 * 1 = 90.Alternatively, 6! / (2! * 2! * 2!) = 90.So, I think 90 is the answer.Wait, but hold on. Let me think about another approach. Suppose we think of it as assigning each class to a teacher, with each teacher getting exactly 2 classes. So, how many ways can we assign 6 distinct classes to 3 distinct teachers, each getting exactly 2 classes.This is equivalent to the number of ways to distribute 6 distinct objects into 3 distinct boxes, each containing exactly 2 objects. The formula for this is 6! / (2! * 2! * 2!) = 90, same as before.Alternatively, using multinomial coefficients: the number is 6! / (2! * 2! * 2!) = 90.So, yes, 90 is the answer.But wait, just to make sure, let me think of a smaller case. Suppose we have 2 teachers and 4 classes, each teacher teaching 2 classes. How many ways?Using the same logic: C(4,2) * C(2,2) = 6 * 1 = 6.Alternatively, 4! / (2! * 2!) = 6.But let's enumerate them:Classes: A, B, C, D.Teachers: T1, T2.Possible assignments:1. T1: A,B; T2: C,D2. T1: A,C; T2: B,D3. T1: A,D; T2: B,C4. T1: B,C; T2: A,D5. T1: B,D; T2: A,C6. T1: C,D; T2: A,BSo, 6 ways, which matches our calculation. So, the formula works for n=2.Similarly, if we have 3 teachers and 6 classes, each teaching 2 classes, the number should be 90, as calculated.Therefore, I think the answer is 90.**Final Answer**The number of different ways is boxed{90}.