💡Alright, so I have this problem about a geometric sequence. Let me see... It says that in the geometric sequence {a_n}, a₁ + a₃ = 5, and a₂ + a₄ = 10. I need to find the value of a₆ + a₈. Hmm, okay. First, I remember that in a geometric sequence, each term is the previous term multiplied by a common ratio, which we usually denote as 'r'. So, the nth term of a geometric sequence can be written as aₙ = a₁ * r^(n-1). That makes sense. So, let me write down what I know:a₁ + a₃ = 5a₂ + a₄ = 10I need to find a₆ + a₈.Let me express each of these terms using the formula for the nth term.a₁ is just a₁.a₃ would be a₁ * r^(3-1) = a₁ * r².Similarly, a₂ is a₁ * r^(2-1) = a₁ * r.a₄ is a₁ * r^(4-1) = a₁ * r³.So, substituting these into the given equations:First equation: a₁ + a₃ = a₁ + a₁ * r² = 5.Second equation: a₂ + a₄ = a₁ * r + a₁ * r³ = 10.Okay, so now I have two equations:1) a₁ + a₁ * r² = 52) a₁ * r + a₁ * r³ = 10Hmm, maybe I can factor out a₁ from both equations to simplify.From equation 1: a₁(1 + r²) = 5From equation 2: a₁ * r(1 + r²) = 10Wait a second, that's interesting. Equation 2 is just equation 1 multiplied by r. Let me check:If I take equation 1: a₁(1 + r²) = 5, and multiply both sides by r, I get a₁ * r(1 + r²) = 5r.But equation 2 is a₁ * r(1 + r²) = 10. So, 5r = 10.Therefore, 5r = 10 implies that r = 10 / 5 = 2.So, the common ratio r is 2. Got it.Now that I know r, I can find a₁ from equation 1.From equation 1: a₁(1 + r²) = 5We know r = 2, so r² = 4.Therefore, a₁(1 + 4) = 5 => a₁ * 5 = 5 => a₁ = 5 / 5 = 1.So, a₁ is 1.Now, I need to find a₆ + a₈.Let me express a₆ and a₈ using the nth term formula.a₆ = a₁ * r^(6-1) = a₁ * r^5a₈ = a₁ * r^(8-1) = a₁ * r^7So, a₆ + a₈ = a₁ * r^5 + a₁ * r^7Factor out a₁ * r^5:a₆ + a₈ = a₁ * r^5 (1 + r²)We already know a₁ = 1 and r = 2, so let's plug those in.First, compute r^5: 2^5 = 32Then, compute 1 + r²: 1 + 4 = 5So, a₆ + a₈ = 1 * 32 * 5 = 160Therefore, the value of a₆ + a₈ is 160.Wait, let me double-check my steps to make sure I didn't make any mistakes.1. Expressed a₃, a₂, a₄ in terms of a₁ and r. That seems correct.2. Factored out a₁ in both equations, leading to equation 1: a₁(1 + r²) = 5 and equation 2: a₁ * r(1 + r²) = 10. That also seems correct.3. Noticing that equation 2 is equation 1 multiplied by r, so 5r = 10, hence r = 2. That makes sense.4. Substituted r = 2 back into equation 1 to find a₁: a₁(1 + 4) = 5 => a₁ = 1. Correct.5. Expressed a₆ and a₈ in terms of a₁ and r, then factored out a₁ * r^5, leading to a₁ * r^5 * (1 + r²). Plugged in the values: 1 * 32 * 5 = 160. That seems right.I think all the steps are correct. So, the answer should be 160.**Final Answer**The value of (a_6 + a_8) is boxed{160}.

💡Alright, so I've got this problem here: find all triples of positive integers (a, b, c) such that a! + b! equals 2 raised to the power of c!. That is,a! + b! = 2^{c!}Hmm, factorials and exponents. Factorials grow really fast, and so do exponentials, especially with factorials in the exponent. So, I think this problem is going to have only a few solutions, maybe even just one or two.First, let me understand the equation better. a! and b! are factorials of positive integers, so they are positive integers as well. Their sum is equal to 2^{c!}, which is also a positive integer. So, all three variables a, b, c are positive integers.Since factorials are involved, maybe I can start by testing small values for a, b, and c. Let's see.Let me consider the case where a and b are both 1. Then, a! + b! = 1! + 1! = 1 + 1 = 2. So, 2 = 2^{c!}. That means c! must be 1 because 2^1 = 2. So, c! = 1 implies c = 1, since 1! = 1. So, (1, 1, 1) is a solution.Okay, that's one solution. Let me see if there are others.What if a = 1 and b = 2? Then, a! + b! = 1! + 2! = 1 + 2 = 3. So, 3 = 2^{c!}. But 3 is not a power of 2, so that doesn't work.What if a = 1 and b = 3? Then, a! + b! = 1! + 3! = 1 + 6 = 7. Again, 7 is not a power of 2. Similarly, a = 1 and b = 4 gives 1 + 24 = 25, which isn't a power of 2. So, it seems like when a = 1 and b > 1, the sum isn't a power of 2.What if a = 2 and b = 2? Then, a! + b! = 2! + 2! = 2 + 2 = 4. So, 4 = 2^{c!}. That means c! must be 2 because 2^2 = 4. So, c! = 2 implies c = 2, since 2! = 2. So, (2, 2, 2) is another solution.Alright, that's another solution. Let me check a = 2 and b = 3. Then, a! + b! = 2! + 3! = 2 + 6 = 8. So, 8 = 2^{c!}. That means c! must be 3 because 2^3 = 8. But c! = 3 implies c = 3, since 3! = 6, which is not equal to 3. Wait, that doesn't make sense. So, c! = 3 would require c to be such that c! = 3, but 3 is not a factorial of any integer except 3 itself, but 3! is 6, not 3. So, actually, there's no c such that c! = 3. Therefore, this case doesn't give a solution.Hmm, so a = 2 and b = 3 doesn't work. What about a = 2 and b = 4? Then, a! + b! = 2! + 4! = 2 + 24 = 26. 26 is not a power of 2, so that doesn't work either.What if a = 3 and b = 3? Then, a! + b! = 6 + 6 = 12. 12 is not a power of 2. Similarly, a = 3 and b = 4 gives 6 + 24 = 30, which isn't a power of 2. So, it seems like when both a and b are 3 or higher, the sum is getting too big and not a power of 2.Wait, let me think about this more systematically. Maybe I can consider different cases based on the values of a and b.Case 1: a = b.In this case, the equation becomes 2 * a! = 2^{c!}. So, 2 * a! = 2^{c!}, which simplifies to a! = 2^{c! - 1}.So, a! must be a power of 2. But factorials are products of consecutive integers, so except for 1! and 2!, they are not powers of 2.Wait, 1! = 1 = 2^0, which is a power of 2, and 2! = 2 = 2^1, which is also a power of 2. 3! = 6, which is not a power of 2, 4! = 24, not a power of 2, and so on. So, the only possible a in this case are a = 1 and a = 2.For a = 1: 1! = 1 = 2^{c! - 1}. So, 2^{c! - 1} = 1 implies c! - 1 = 0, so c! = 1. Therefore, c = 1.For a = 2: 2! = 2 = 2^{c! - 1}. So, 2^{c! - 1} = 2 implies c! - 1 = 1, so c! = 2. Therefore, c = 2.So, in the case where a = b, the only solutions are (1, 1, 1) and (2, 2, 2).Case 2: a ≠ b.Without loss of generality, assume a < b.Then, a! + b! = a!(1 + b!/a!) = a!(1 + (b)(b-1)...(a+1)).Since a < b, b! is divisible by a!, so b!/a! is an integer greater than 1.So, the equation becomes a! * (1 + k) = 2^{c!}, where k is an integer greater than 1.Therefore, a! must be a power of 2, and (1 + k) must also be a power of 2.From earlier, we saw that a! is a power of 2 only when a = 1 or a = 2.So, let's consider a = 1.Then, a! = 1, so the equation becomes 1 + b! = 2^{c!}.So, 1 + b! must be a power of 2.Let's check for small values of b.b = 1: 1 + 1 = 2, which is 2^1. So, c! = 1, c = 1. So, (1, 1, 1) is a solution, which we already have.b = 2: 1 + 2 = 3, which is not a power of 2.b = 3: 1 + 6 = 7, not a power of 2.b = 4: 1 + 24 = 25, not a power of 2.b = 5: 1 + 120 = 121, not a power of 2.b = 6: 1 + 720 = 721, not a power of 2.And it's clear that as b increases, b! grows very rapidly, and 1 + b! will not be a power of 2 because it's odd and greater than 2, and powers of 2 are even except for 2^0 = 1.Wait, actually, 1 + b! is even only when b! is odd, which happens only when b = 1, because 1! = 1, which is odd. For b >= 2, b! is even, so 1 + b! is odd. But 2^{c!} is even unless c! = 0, which is not possible because c is a positive integer. So, 2^{c!} is always even for c >= 1. Therefore, 1 + b! is odd, and 2^{c!} is even, so they can't be equal unless 1 + b! is even, which only happens when b = 1. But in that case, we already have the solution (1,1,1).Therefore, for a = 1 and b > 1, there are no solutions.Now, let's consider a = 2.Then, a! = 2, so the equation becomes 2 + b! = 2^{c!}.So, 2 + b! must be a power of 2.Let's check for small values of b.b = 2: 2 + 2 = 4 = 2^2. So, c! = 2, which implies c = 2. So, (2, 2, 2) is a solution, which we already have.b = 3: 2 + 6 = 8 = 2^3. So, c! = 3. But c! = 3 implies c = 3, but 3! = 6, which is not equal to 3. So, no solution here.b = 4: 2 + 24 = 26, which is not a power of 2.b = 5: 2 + 120 = 122, not a power of 2.b = 6: 2 + 720 = 722, not a power of 2.Again, as b increases, b! grows rapidly, and 2 + b! is not a power of 2.Wait, let me think about this modulo something. Maybe modulo 4.For b >= 4, b! is divisible by 4, because 4! = 24, and higher factorials include 4 as a factor. So, b! ≡ 0 mod 4 for b >= 4.Therefore, 2 + b! ≡ 2 mod 4.But 2^{c!} is equal to 2 mod 4 only when c! = 1, because 2^1 = 2, and 2^{c!} for c! >= 2 is 0 mod 4.Wait, 2^{c!} mod 4:If c! = 1, 2^1 = 2 mod 4.If c! >= 2, 2^{c!} is divisible by 4, so 0 mod 4.So, 2 + b! ≡ 2 mod 4, which would require 2^{c!} ≡ 2 mod 4, which only happens when c! = 1, i.e., c = 1.But if c = 1, then 2^{c!} = 2^1 = 2. So, 2 + b! = 2 implies b! = 0, which is impossible because b is a positive integer. Therefore, for b >= 4, there's no solution.So, the only possible b for a = 2 is b = 2 or b = 3.We already saw that b = 2 gives a solution, and b = 3 doesn't because c! would have to be 3, which isn't possible.Therefore, for a = 2 and b > 2, there are no solutions.Case 3: a >= 3.If a >= 3, then a! is divisible by 3, and b! is also divisible by 3 if b >= 3.So, a! + b! is divisible by 3.But 2^{c!} is a power of 2, which is never divisible by 3. Therefore, a! + b! cannot be equal to 2^{c!} if both a and b are >= 3.Therefore, the only possible solutions are when a and b are both 1 or both 2, which we have already found.So, putting it all together, the only triples (a, b, c) that satisfy the equation are (1, 1, 1) and (2, 2, 2).**Final Answer**The solutions are boxed{(1, 1, 1)} and boxed{(2, 2, 2)}.

💡Alright, so I have this problem about a store buying two types of eye-protection lamps, A and B. I need to figure out the cost price of each lamp and then determine how to maximize profit based on some constraints. Let me try to break this down step by step.First, part (1) asks for the cost price of each lamp. It says that the cost price of lamp A is 40 yuan more than lamp B. Okay, so if I let the cost price of lamp B be, say, x yuan, then lamp A would be x + 40 yuan. That makes sense.Next, it mentions that the quantity of type A lamps that can be purchased with 2000 yuan is the same as the quantity of type B lamps that can be purchased with 1600 yuan. Hmm, so if I have 2000 yuan and I buy lamp A, the number I can get is the same as if I have 1600 yuan and buy lamp B.Let me write that as an equation. The number of lamps you can buy is the total money divided by the cost per lamp. So, for lamp A, it's 2000 divided by (x + 40), and for lamp B, it's 1600 divided by x. And these two quantities are equal.So, 2000 / (x + 40) = 1600 / x.Now, I need to solve for x. Let me cross-multiply to get rid of the fractions.2000 * x = 1600 * (x + 40).Expanding the right side: 1600x + 1600*40 = 1600x + 64000.So, 2000x = 1600x + 64000.Subtract 1600x from both sides: 400x = 64000.Divide both sides by 400: x = 64000 / 400 = 160.So, x is 160 yuan. That means lamp B costs 160 yuan, and lamp A, which is 40 yuan more, is 160 + 40 = 200 yuan.Okay, that seems straightforward. So, part (1) is done: lamp A is 200 yuan, lamp B is 160 yuan.Now, moving on to part (2). The store plans to purchase a total of 80 lamps with a budget not exceeding 14550 yuan. The selling prices are 300 yuan for A and 200 yuan for B. All lamps are sold, and we need to maximize profit.First, let's understand what's given:- Total lamps: 80 (so, if I buy 'a' number of A lamps, then B lamps would be 80 - a).- Total budget: 14550 yuan. So, the cost of A lamps plus the cost of B lamps should be less than or equal to 14550.- Selling prices: A is sold for 300, B for 200.So, profit is calculated as (Selling Price - Cost Price) for each lamp, multiplied by the number of lamps sold.So, for lamp A: Profit per lamp = 300 - 200 = 100 yuan.For lamp B: Profit per lamp = 200 - 160 = 40 yuan.Therefore, total profit y = 100a + 40(80 - a).But we also have the budget constraint: 200a + 160(80 - a) ≤ 14550.Let me write that down:200a + 160(80 - a) ≤ 14550.Let me simplify this inequality.First, expand the left side:200a + 160*80 - 160a ≤ 14550.Calculate 160*80: 160*80 = 12800.So, 200a + 12800 - 160a ≤ 14550.Combine like terms: (200a - 160a) + 12800 ≤ 14550.That's 40a + 12800 ≤ 14550.Subtract 12800 from both sides: 40a ≤ 14550 - 12800.Calculate 14550 - 12800: 14550 - 12800 = 1750.So, 40a ≤ 1750.Divide both sides by 40: a ≤ 1750 / 40.Calculate 1750 / 40: 1750 ÷ 40 = 43.75.But since a must be an integer (you can't buy a fraction of a lamp), the maximum number of A lamps we can buy is 43.So, a ≤ 43.Now, since we want to maximize profit, and the profit per A lamp is higher than that of B (100 vs. 40), we should buy as many A lamps as possible within the budget constraint.Therefore, buying 43 A lamps and 80 - 43 = 37 B lamps.Now, let's calculate the total profit.Profit y = 100a + 40(80 - a).Substitute a = 43:y = 100*43 + 40*(80 - 43).Calculate 100*43: 4300.Calculate 80 - 43: 37.Calculate 40*37: 1480.So, total profit y = 4300 + 1480 = 5780 yuan.Let me double-check the budget to ensure we don't exceed it.Cost of 43 A lamps: 43*200 = 8600 yuan.Cost of 37 B lamps: 37*160 = 5920 yuan.Total cost: 8600 + 5920 = 14520 yuan.Which is within the 14550 yuan budget.So, everything checks out.Therefore, the store should purchase 43 lamps of type A and 37 lamps of type B to maximize profit, resulting in a maximum profit of 5780 yuan.

💡Okay, so I have this function f(x) defined piecewise. For x less than or equal to 0, it's a quadratic function x squared plus b times x plus c. And for x greater than 0, it's just 2. The problem gives me two specific values: f(-4) equals 2 and f(-2) equals -2. I need to find the number of solutions to the equation f(x) equals x. The options are 1, 2, 3, or 4.Alright, let me start by understanding what's given. The function is piecewise, so it behaves differently depending on whether x is positive or negative. For x ≤ 0, it's a quadratic, which is a parabola. For x > 0, it's a constant function, always equal to 2. They give me two points on this function: when x is -4, f(-4) is 2, and when x is -2, f(-2) is -2. Since both -4 and -2 are less than or equal to 0, these points lie on the quadratic part of the function. So I can use these two points to set up equations and solve for the coefficients b and c.Let me write down the equations. For x = -4, f(-4) = (-4)^2 + b*(-4) + c = 2. That simplifies to 16 - 4b + c = 2. Similarly, for x = -2, f(-2) = (-2)^2 + b*(-2) + c = -2, which simplifies to 4 - 2b + c = -2.So now I have a system of two equations:1. 16 - 4b + c = 22. 4 - 2b + c = -2I need to solve this system for b and c. Let me subtract the second equation from the first to eliminate c. First equation: 16 - 4b + c = 2Second equation: 4 - 2b + c = -2Subtracting the second from the first:(16 - 4b + c) - (4 - 2b + c) = 2 - (-2)16 - 4b + c - 4 + 2b - c = 4Simplify:(16 - 4) + (-4b + 2b) + (c - c) = 412 - 2b = 4Now, solve for b:12 - 2b = 4Subtract 12 from both sides:-2b = -8Divide both sides by -2:b = 4Okay, so b is 4. Now plug this back into one of the original equations to find c. Let me use the second equation because the numbers are smaller.Second equation: 4 - 2b + c = -2Plug in b = 4:4 - 2*(4) + c = -24 - 8 + c = -2-4 + c = -2Add 4 to both sides:c = 2So now I have b = 4 and c = 2. Therefore, the quadratic part of the function is x squared plus 4x plus 2. So the function f(x) is:f(x) = { x² + 4x + 2, if x ≤ 0 { 2, if x > 0Now, I need to solve f(x) = x. That means I have to solve this equation in two parts: one for x ≤ 0 and one for x > 0.First, let's consider x > 0. In this case, f(x) is 2, so the equation becomes 2 = x. So x = 2. But wait, x has to be greater than 0, and 2 is greater than 0, so this is a valid solution. So that's one solution: x = 2.Now, for x ≤ 0, f(x) is the quadratic x² + 4x + 2. So the equation becomes x² + 4x + 2 = x. Let me rearrange this equation to bring all terms to one side:x² + 4x + 2 - x = 0Simplify:x² + 3x + 2 = 0Now, I need to solve this quadratic equation. Let's factor it:x² + 3x + 2 = (x + 1)(x + 2) = 0So the solutions are x = -1 and x = -2. Now, I need to check if these solutions satisfy the condition x ≤ 0. Both -1 and -2 are less than or equal to 0, so both are valid solutions.So from the quadratic part, I have two solutions: x = -1 and x = -2.Adding the solution from the x > 0 part, which is x = 2, I have a total of three solutions: x = -2, x = -1, and x = 2.Wait, but let me double-check if x = 2 is indeed a solution. For x > 0, f(x) is 2, so f(2) = 2. And the equation is f(x) = x, so 2 = 2. That's correct. So x = 2 is a valid solution.Similarly, let me check x = -1. Plugging into f(x): f(-1) = (-1)^2 + 4*(-1) + 2 = 1 - 4 + 2 = -1. So f(-1) = -1, which equals x, so that's correct.For x = -2: f(-2) = (-2)^2 + 4*(-2) + 2 = 4 - 8 + 2 = -2. So f(-2) = -2, which equals x, so that's correct as well.Therefore, all three solutions are valid.Wait, but hold on a second. The quadratic equation x² + 3x + 2 = 0 has two real roots, both negative, so they are within the domain x ≤ 0. So that's two solutions. Then, the equation for x > 0 gives one solution. So in total, three solutions.But let me think again. The function f(x) is a piecewise function. For x ≤ 0, it's a quadratic, and for x > 0, it's a constant function. So when solving f(x) = x, we have to consider both parts.For x > 0, f(x) = 2, so 2 = x, which is x = 2. That's straightforward.For x ≤ 0, f(x) is quadratic, so solving x² + 4x + 2 = x gives x² + 3x + 2 = 0, which factors to (x + 1)(x + 2) = 0, giving x = -1 and x = -2. Both are valid because they are less than or equal to 0.So that's three solutions in total. So the answer should be 3, which is option C.But wait, let me visualize the function to make sure I haven't missed anything. For x ≤ 0, it's a quadratic opening upwards because the coefficient of x² is positive. The vertex of this quadratic is at x = -b/(2a) = -4/(2*1) = -2. So the vertex is at x = -2, and f(-2) = -2, which matches the given point. So the parabola has its vertex at (-2, -2) and it opens upwards.So the quadratic part starts at x = 0, where f(0) = 0 + 0 + 2 = 2. So at x = 0, the function is 2, and for x > 0, it's also 2. So the function is continuous at x = 0.Now, when solving f(x) = x, for x > 0, it's just x = 2, which is straightforward.For x ≤ 0, the quadratic x² + 4x + 2 intersects the line y = x at x = -1 and x = -2. So that's two points.Therefore, in total, three solutions: x = -2, x = -1, and x = 2.Wait, but let me check if the quadratic actually intersects y = x at two points. Since the quadratic is opening upwards, and it's vertex is at (-2, -2), which is below the line y = x at that point (since y = x at x = -2 is y = -2, so they meet there). Then, as x increases towards 0, the quadratic goes from (-2, -2) up to (0, 2). The line y = x at x = 0 is y = 0, so the quadratic is at 2, which is above y = x. So the quadratic starts at (-2, -2), goes up, crosses y = x at x = -1, and then continues to (0, 2). So yes, it intersects y = x twice: once at x = -2 and again at x = -1.Wait, but hold on. At x = -2, the quadratic is at y = -2, which is equal to x, so that's one intersection. Then, as x increases, the quadratic goes up, crossing y = x again at x = -1. So that's two intersections on the quadratic side, and one on the constant side, making three solutions in total.Therefore, the number of solutions is 3, which is option C.I think that's thorough. I considered both parts of the piecewise function, solved the equations, checked the validity of the solutions, and even visualized the function to make sure. So I'm confident the answer is 3.**Final Answer**boxed{C}

💡Okay, so I need to figure out how much Tom will have to pay in total for his trip to Barbados. Let me start by listing out all the expenses he has mentioned.First, there are the medical expenses: he needs 10 different vaccines, each costing 45, and a doctor's visit that costs 250. So, the total medical cost before insurance would be the cost of the vaccines plus the doctor's visit. Let me calculate that:10 vaccines * 45 per vaccine = 450450 (vaccines) + 250 (doctor's visit) = 700Now, insurance covers 80% of these medical bills. So, I need to find out how much the insurance will cover and how much Tom will have to pay out of pocket.Insurance coverage: 80% of 700 = 0.80 * 700 = 560Tom's out-of-pocket medical cost: 700 - 560 = 140Okay, so Tom will have to pay 140 for his medical expenses.Next, the round-trip flight costs 1200. That seems straightforward.Now, Tom also needs to pay for lodging, transportation, and food. These are local expenses in Barbados, so they need to be paid in Barbados Dollars (BBD). The exchange rate is 1 USD = 2 BBD.Let's calculate each of these local expenses:Lodging: 7 nights * 150 per night = 1050Transportation: 200Food: 7 days * 60 per day = 420Total local expenses in USD: 1050 + 200 + 420 = 1670Now, converting this total to BBD:1670 * 2 BBD/USD = 3340 BBDBut Tom's bank charges a 3% fee for currency conversion. So, I need to calculate the conversion fee and add it to the total expenses.Conversion fee: 3% of 3340 BBD = 0.03 * 3340 = 100.2 BBDNow, converting this fee back to USD to add it to Tom's total expenses:100.2 BBD / 2 BBD/USD = 50.10So, the total expenses in USD are:Medical cost (out-of-pocket): 140Flight cost: 1200Local expenses (lodging, transportation, food): 1670Currency conversion fee: 50.10Adding them all up:140 + 1200 + 1670 + 50.10 = 3060.10Therefore, Tom will have to pay a total of 3060.10 for his trip to Barbados.

💡Okay, so I have this probability problem here: Four cards are dealt at random from a standard deck of 52 cards. What is the probability that the first card is a heart, the second card is a King, the third card is a King, and the fourth card is an Ace?Hmm, let's see. I need to figure out the probability of this specific sequence happening. I think I'll need to break it down step by step, considering the probabilities at each stage and how each draw affects the next.First, the first card needs to be a heart. There are 13 hearts in a standard deck of 52 cards. So, the probability that the first card is a heart is 13/52, which simplifies to 1/4. But wait, I need to consider if the first heart could be a King or an Ace because that might affect the probabilities of the subsequent draws. Hmm, so maybe I need to break this down into different cases.Case 1: The first card is a heart that is neither a King nor an Ace. There are 11 such hearts (since there are 13 hearts total, minus the King and Ace of hearts). So, the probability here is 11/52.Case 2: The first card is the King of hearts. There's only one King of hearts, so the probability is 1/52.Case 3: The first card is the Ace of hearts. Similarly, there's only one Ace of hearts, so the probability is 1/52.Okay, so these are the three possible cases for the first card. Now, I need to consider each case separately and then sum up the probabilities because these are mutually exclusive events.Starting with Case 1: First card is a heart that's neither a King nor an Ace.After drawing a heart that's not a King or Ace, there are now 51 cards left in the deck. The second card needs to be a King. There are four Kings in the deck, and none have been removed yet because the first card wasn't a King. So, the probability of drawing a King next is 4/51.Now, after drawing a King, there are 50 cards left, and we need to draw another King. Since one King has already been drawn, there are three Kings left. So, the probability here is 3/50.Finally, the fourth card needs to be an Ace. There are four Aces in the deck, and none have been drawn yet because the first card was a heart that wasn't an Ace, and the next two were Kings. So, the probability of drawing an Ace is 4/49.So, for Case 1, the probability is:(11/52) * (4/51) * (3/50) * (4/49)Let me calculate that:11 * 4 = 4444 * 3 = 132132 * 4 = 528Denominator: 52 * 51 = 26522652 * 50 = 132600132600 * 49 = 6497400So, 528 / 6497400. Let me simplify that. Dividing numerator and denominator by 12:528 ÷ 12 = 446497400 ÷ 12 = 541450So, 44 / 541450. Hmm, that's approximately 0.00008127, or about 1 in 12317.Okay, that's Case 1.Now, moving on to Case 2: The first card is the King of hearts.Probability of drawing the King of hearts first is 1/52.After drawing the King of hearts, there are 51 cards left. Now, the second card needs to be a King. But since we've already drawn one King (the King of hearts), there are only three Kings left. So, the probability is 3/51.Then, the third card also needs to be a King. After drawing two Kings, there are two Kings left out of 50 cards. So, the probability is 2/50.Finally, the fourth card needs to be an Ace. There are four Aces in the deck, and none have been drawn yet because the first card was a King, and the next two were Kings. So, the probability is 4/49.So, for Case 2, the probability is:(1/52) * (3/51) * (2/50) * (4/49)Calculating that:1 * 3 = 33 * 2 = 66 * 4 = 24Denominator: 52 * 51 = 26522652 * 50 = 132600132600 * 49 = 6497400So, 24 / 6497400. Simplifying:24 / 6497400 = 2 / 541450 ≈ 0.0000037, or about 1 in 270,000.That's much smaller than Case 1.Now, Case 3: The first card is the Ace of hearts.Probability of drawing the Ace of hearts first is 1/52.After drawing the Ace of hearts, there are 51 cards left. The second card needs to be a King. There are four Kings in the deck, none of which have been drawn yet because the first card was an Ace. So, the probability is 4/51.Then, the third card also needs to be a King. After drawing one King, there are three left out of 50 cards. So, the probability is 3/50.Finally, the fourth card needs to be an Ace. Since we've already drawn one Ace (the Ace of hearts), there are three Aces left out of 49 cards. So, the probability is 3/49.So, for Case 3, the probability is:(1/52) * (4/51) * (3/50) * (3/49)Calculating that:1 * 4 = 44 * 3 = 1212 * 3 = 36Denominator: 52 * 51 = 26522652 * 50 = 132600132600 * 49 = 6497400So, 36 / 6497400. Simplifying:36 / 6497400 = 3 / 541450 ≈ 0.00000554, or about 1 in 180,000.Again, much smaller than Case 1.Now, to find the total probability, I need to add up the probabilities of all three cases because they are mutually exclusive.So, total probability = Case 1 + Case 2 + Case 3Which is:528 / 6497400 + 24 / 6497400 + 36 / 6497400Adding the numerators:528 + 24 = 552552 + 36 = 588So, total probability is 588 / 6497400Simplify that:Divide numerator and denominator by 12:588 ÷ 12 = 496497400 ÷ 12 = 541450So, 49 / 541450Simplify further by dividing numerator and denominator by 7:49 ÷ 7 = 7541450 ÷ 7 = 77350So, 7 / 77350Simplify further by dividing numerator and denominator by 7 again:7 ÷ 7 = 177350 ÷ 7 = 11050So, 1 / 11050Wait, that doesn't seem right because earlier, Case 1 was 528 / 6497400 which simplified to 44 / 541450, and adding the other cases gave us 588 / 6497400 which simplifies to 1 / 11050.But earlier, I thought Case 1 was approximately 1 / 12317, but now the total is 1 / 11050, which is slightly higher. That makes sense because we added the smaller probabilities from Cases 2 and 3.But let me double-check my calculations to make sure I didn't make a mistake.Starting with total probability:Case 1: 11/52 * 4/51 * 3/50 * 4/49 = (11*4*3*4)/(52*51*50*49) = 528 / 6497400Case 2: 1/52 * 3/51 * 2/50 * 4/49 = (1*3*2*4)/(52*51*50*49) = 24 / 6497400Case 3: 1/52 * 4/51 * 3/50 * 3/49 = (1*4*3*3)/(52*51*50*49) = 36 / 6497400Adding them up: 528 + 24 + 36 = 588So, 588 / 6497400Simplify:Divide numerator and denominator by 12: 588 ÷ 12 = 49, 6497400 ÷ 12 = 541450So, 49 / 541450Divide numerator and denominator by 7: 49 ÷ 7 = 7, 541450 ÷ 7 = 77350So, 7 / 77350Divide numerator and denominator by 7 again: 7 ÷ 7 = 1, 77350 ÷ 7 = 11050So, 1 / 11050Wait, but earlier I thought the answer was 1 / 12317, but that was just for Case 1. The total probability is 1 / 11050.But let me check if I considered all cases correctly.Yes, I considered three cases: first card is a heart that's neither King nor Ace, first card is King of hearts, and first card is Ace of hearts.I think that's comprehensive because any heart could be either a regular heart, the King of hearts, or the Ace of hearts.So, adding up all three cases gives the total probability.Therefore, the final probability is 1 / 11050.But wait, let me check if that's correct.Alternatively, maybe I can approach this problem without splitting into cases.Let's see. The probability that the first card is a heart, the second is a King, the third is a King, and the fourth is an Ace.But the problem is that some of these events can overlap. For example, the first heart could be the King of hearts or the Ace of hearts, which affects the counts for the subsequent draws.So, perhaps the initial approach of splitting into cases is necessary.Alternatively, maybe I can use permutations.Total number of ways to draw four cards: 52 * 51 * 50 * 49Number of favorable outcomes:First card: heart (13 options)Second card: King (4 options)Third card: King (3 options remaining)Fourth card: Ace (4 options)But wait, this approach doesn't account for overlaps where the first heart could be a King or Ace, which would reduce the number of available Kings or Aces for the subsequent draws.So, to avoid overcounting, it's better to split into cases.Therefore, the initial approach is correct.So, the total probability is 1 / 11050.But let me check if that's accurate.Alternatively, maybe I can calculate it as:Probability = (Number of favorable sequences) / (Total number of possible sequences)Total number of possible sequences: 52 * 51 * 50 * 49Number of favorable sequences:Case 1: First card is a heart that's neither King nor Ace (11 options), second card is a King (4 options), third card is a King (3 options), fourth card is an Ace (4 options). So, 11 * 4 * 3 * 4Case 2: First card is King of hearts (1 option), second card is a King (3 options), third card is a King (2 options), fourth card is an Ace (4 options). So, 1 * 3 * 2 * 4Case 3: First card is Ace of hearts (1 option), second card is a King (4 options), third card is a King (3 options), fourth card is an Ace (3 options). So, 1 * 4 * 3 * 3Total favorable sequences: 11*4*3*4 + 1*3*2*4 + 1*4*3*3 = 528 + 24 + 36 = 588So, probability = 588 / (52*51*50*49) = 588 / 6497400 = 1 / 11050Yes, that matches my earlier calculation.Therefore, the probability is 1 / 11050.