answer：First, I need to determine the total number of files Amy initially has on her flash drive. She has 4.0 music files and 21.0 video files.Next, I'll add these two quantities together to find the initial total number of files:4.0 (music files) + 21.0 (video files) = 25.0 filesAmy then downloads an additional 23.0 picture files. To find the new total number of files on her flash drive, I'll add the downloaded picture files to the initial total:25.0 (existing files) + 23.0 (picture files) = 48.0 filesTherefore, the total number of files on Amy's flash drive after downloading the picture files is 48.0.

answer：Okay, so I have this problem about transforming the graph of a function and then solving an equation involving it. Let me try to break it down step by step.First, part (1) asks for the analytical expression of the function f(x) and the equation of its axis of symmetry. The function g(x) is given as cos x. The transformations applied are: first, stretching the vertical coordinates by a factor of 2, and then shifting the graph to the right by π/2 units.Alright, stretching vertically by a factor of 2 would mean multiplying the function by 2. So, if g(x) = cos x, then after stretching, it becomes 2 cos x. Then, shifting it to the right by π/2 units. In function transformations, shifting to the right by c units is done by replacing x with (x - c). So, replacing x with (x - π/2) in 2 cos x gives 2 cos(x - π/2).Hmm, but wait, cos(x - π/2) is equal to sin x because cos(θ - π/2) = sin θ. So, 2 cos(x - π/2) simplifies to 2 sin x. So, f(x) = 2 sin x.Now, the axis of symmetry. For the sine function, which is f(x) = 2 sin x, the graph is a sine wave. The sine function doesn't have vertical lines of symmetry like a parabola does, but it does have points of symmetry. However, if we're talking about an axis of symmetry, maybe it's referring to the vertical lines where the function reaches its maximum or minimum.The standard sine function, sin x, has its maximum at π/2, 5π/2, etc., and minimum at 3π/2, 7π/2, etc. So, the axis of symmetry would be the vertical lines passing through these points. So, for f(x) = 2 sin x, the axis of symmetry would be x = π/2 + kπ, where k is any integer. That makes sense because each peak and trough is separated by π units.So, part (1) seems manageable. I think f(x) is 2 sin x, and the axis of symmetry is x = π/2 + kπ for integers k.Moving on to part (2). It says that the equation f(x) + g(x) = m has two different solutions α and β in the interval [0, 2π]. Then, it has two subparts: ① finding the range of m, and ② proving that cos(α - β) equals (2m²)/5 - 1.Let me tackle ① first. So, f(x) is 2 sin x, and g(x) is cos x. So, the equation is 2 sin x + cos x = m.I need to find the range of m such that this equation has two different solutions in [0, 2π]. Hmm, this seems like it can be approached by combining the sine and cosine terms into a single trigonometric function.I remember that expressions like a sin x + b cos x can be rewritten as R sin(x + φ) or R cos(x + φ), where R is the amplitude and φ is the phase shift. Let me try that.So, 2 sin x + cos x can be written as R sin(x + φ). Let's compute R and φ.The formula for R is sqrt(a² + b²), where a is the coefficient of sin x and b is the coefficient of cos x. So here, a = 2 and b = 1, so R = sqrt(2² + 1²) = sqrt(4 + 1) = sqrt(5).Then, φ is given by tan φ = b/a, but wait, actually, it's tan φ = b/a if we're using the form R sin(x + φ). Wait, let me recall the exact formula.Actually, when expressing a sin x + b cos x as R sin(x + φ), we have:a = R cos φb = R sin φSo, tan φ = b/a. So, tan φ = 1/2. Therefore, φ = arctan(1/2). So, φ is some angle whose tangent is 1/2.Therefore, 2 sin x + cos x = sqrt(5) sin(x + φ), where φ = arctan(1/2). So, the equation becomes sqrt(5) sin(x + φ) = m.So, sin(x + φ) = m / sqrt(5). For this equation to have solutions, the absolute value of m / sqrt(5) must be less than or equal to 1. So, |m| / sqrt(5) ≤ 1, which implies |m| ≤ sqrt(5). So, m must be in the interval [-sqrt(5), sqrt(5)].But the problem says that the equation has two different solutions in [0, 2π]. So, when does sin(x + φ) = k have two solutions in an interval?Well, for the equation sin θ = k, where θ is in [φ, 2π + φ], which is just a shifted interval, the number of solutions depends on k. If |k| < 1, there are two solutions in [0, 2π]. If |k| = 1, there is one solution. So, in our case, sin(x + φ) = m / sqrt(5) will have two solutions when |m / sqrt(5)| < 1, which is |m| < sqrt(5). If |m| = sqrt(5), then sin(x + φ) = ±1, which would have only one solution each in [0, 2π]. So, to have two different solutions, m must be in (-sqrt(5), sqrt(5)).So, the range of m is (-sqrt(5), sqrt(5)). That should be the answer for part ①.Now, part ② asks to prove that cos(α - β) = (2m²)/5 - 1.Hmm, okay. So, α and β are two different solutions in [0, 2π] of the equation 2 sin x + cos x = m. So, they satisfy 2 sin α + cos α = m and 2 sin β + cos β = m.We need to find cos(α - β). Hmm. Let me think about how to relate α and β.Since both α and β satisfy the equation 2 sin x + cos x = m, they are two different solutions. So, perhaps we can write equations for both and then find a relationship between α and β.Alternatively, since we have expressed 2 sin x + cos x as sqrt(5) sin(x + φ), where φ = arctan(1/2), then the equation becomes sqrt(5) sin(x + φ) = m. So, sin(x + φ) = m / sqrt(5).So, the solutions are x + φ = arcsin(m / sqrt(5)) and x + φ = π - arcsin(m / sqrt(5)), because sine is positive in the first and second quadrants.Therefore, the solutions are:x = arcsin(m / sqrt(5)) - φandx = π - arcsin(m / sqrt(5)) - φ.So, let me denote θ = arcsin(m / sqrt(5)). Then, the two solutions are x = θ - φ and x = π - θ - φ.Therefore, α = θ - φ and β = π - θ - φ.So, α - β = (θ - φ) - (π - θ - φ) = θ - φ - π + θ + φ = 2θ - π.Therefore, cos(α - β) = cos(2θ - π). Hmm, cos(2θ - π) can be simplified.We know that cos(2θ - π) = cos(π - 2θ) because cosine is even. And cos(π - 2θ) = -cos(2θ). So, cos(α - β) = -cos(2θ).Now, θ is arcsin(m / sqrt(5)). So, let's compute cos(2θ). Using the double-angle formula: cos(2θ) = 1 - 2 sin²θ.Since θ = arcsin(m / sqrt(5)), sin θ = m / sqrt(5). Therefore, cos(2θ) = 1 - 2(m² / 5).Therefore, cos(α - β) = -cos(2θ) = -[1 - 2(m² / 5)] = -1 + (2m²)/5 = (2m²)/5 - 1.So, that's the proof. Therefore, cos(α - β) equals (2m²)/5 - 1.Wait, let me double-check that step where I said cos(π - 2θ) = -cos(2θ). Yes, because cos(π - x) = -cos x. So, cos(π - 2θ) = -cos(2θ). So, that part is correct.Also, the double-angle formula: cos(2θ) = 1 - 2 sin²θ. That's correct. So, substituting sin θ = m / sqrt(5), we get cos(2θ) = 1 - 2(m² / 5). So, that's correct.Therefore, putting it all together, cos(α - β) = -cos(2θ) = -[1 - 2(m² / 5)] = (2m²)/5 - 1. So, that's the result.I think that makes sense. So, the key steps were expressing the equation as a single sine function, finding the two solutions, relating α and β, and then using trigonometric identities to find cos(α - β).Let me just recap to make sure I didn't miss anything.1. Transformed g(x) = cos x by vertical stretch and horizontal shift to get f(x) = 2 sin x.2. For part (2), combined f(x) + g(x) into a single sine function with amplitude sqrt(5).3. Found the range of m by considering the amplitude.4. For part ②, expressed the solutions α and β in terms of arcsin and then found the difference α - β.5. Used trigonometric identities to express cos(α - β) in terms of m.Everything seems to check out. I think I've got it.**Final Answer**(1) The function is boxed{2sin x} and its axis of symmetry is x = kpi + frac{pi}{2} for any integer k.(2) ① The range of m is boxed{(-sqrt{5}, sqrt{5})}.② cos(alpha - beta) = boxed{frac{2m^2}{5} - 1}.

answer：Okay, so I need to find positive real numbers a, b, and c that maximize the product abc, given the constraint b(a² + 2) + c(a + 2) = 12. Hmm, this seems like an optimization problem with a constraint. I remember that for such problems, methods like Lagrange multipliers can be useful. But I'm not entirely sure how to apply them here, so maybe I should start by understanding the problem better.First, let me write down what I need to maximize and the constraint:Maximize: abcSubject to: b(a² + 2) + c(a + 2) = 12All variables a, b, c are positive real numbers.Alright, so I have three variables and one equation. That means I can express two variables in terms of the third, but since I need to maximize a product, maybe I can reduce the problem to a single variable.Let me think about expressing b and c in terms of a. From the constraint:b(a² + 2) + c(a + 2) = 12I can solve for b and c. Let's see, if I solve for b:b = (12 - c(a + 2)) / (a² + 2)Similarly, solving for c:c = (12 - b(a² + 2)) / (a + 2)But this might complicate things because both b and c are expressed in terms of each other. Maybe it's better to express one variable in terms of the other and substitute back into the product abc.Alternatively, I can use the method of Lagrange multipliers. I think that's a more systematic approach for optimization with constraints.So, let's recall how Lagrange multipliers work. If I have a function f(a, b, c) to maximize subject to a constraint g(a, b, c) = 0, then I can set up the Lagrangian:L(a, b, c, λ) = abc + λ(12 - b(a² + 2) - c(a + 2))Then, I need to take partial derivatives of L with respect to a, b, c, and λ, and set them equal to zero.Let me compute each partial derivative:1. Partial derivative with respect to a:∂L/∂a = bc + λ(-2ab - c) = 02. Partial derivative with respect to b:∂L/∂b = ac + λ(-a² - 2) = 03. Partial derivative with respect to c:∂L/∂c = ab + λ(-a - 2) = 04. Partial derivative with respect to λ:∂L/∂λ = 12 - b(a² + 2) - c(a + 2) = 0So, now I have four equations:1. bc - λ(2ab + c) = 02. ac - λ(a² + 2) = 03. ab - λ(a + 2) = 04. b(a² + 2) + c(a + 2) = 12Hmm, okay. So, I need to solve this system of equations. Let's see if I can express λ from equations 2 and 3 and then equate them.From equation 2:ac = λ(a² + 2) => λ = ac / (a² + 2)From equation 3:ab = λ(a + 2) => λ = ab / (a + 2)So, setting these two expressions for λ equal:ac / (a² + 2) = ab / (a + 2)Let me simplify this equation. Multiply both sides by (a² + 2)(a + 2):ac(a + 2) = ab(a² + 2)Divide both sides by a (since a is positive, we can do this):c(a + 2) = b(a² + 2)Interesting, so c(a + 2) = b(a² + 2). Let me note this as equation 5.Now, let's look back at the constraint equation 4:b(a² + 2) + c(a + 2) = 12But from equation 5, c(a + 2) = b(a² + 2). So, substituting into equation 4:b(a² + 2) + b(a² + 2) = 12So, 2b(a² + 2) = 12 => b(a² + 2) = 6 => b = 6 / (a² + 2)Great, so now I can express b in terms of a. Similarly, from equation 5:c(a + 2) = b(a² + 2) = 6So, c = 6 / (a + 2)Now, I have expressions for both b and c in terms of a:b = 6 / (a² + 2)c = 6 / (a + 2)Now, let's substitute these into the product abc:abc = a * (6 / (a² + 2)) * (6 / (a + 2)) = (36a) / [(a² + 2)(a + 2)]So, now the problem reduces to maximizing the function f(a) = (36a) / [(a² + 2)(a + 2)] for a > 0.To find the maximum, I can take the derivative of f(a) with respect to a, set it equal to zero, and solve for a.Let me compute f'(a):First, let me write f(a) as:f(a) = 36a / [(a² + 2)(a + 2)] = 36a / (a³ + 2a² + 2a + 4)Let me denote the denominator as D(a) = a³ + 2a² + 2a + 4Then, f(a) = 36a / D(a)Using the quotient rule, f'(a) = [36D(a) - 36a D'(a)] / [D(a)]²Compute D'(a):D'(a) = 3a² + 4a + 2So, f'(a) = [36(a³ + 2a² + 2a + 4) - 36a(3a² + 4a + 2)] / (a³ + 2a² + 2a + 4)²Let me factor out 36:f'(a) = 36 [ (a³ + 2a² + 2a + 4) - a(3a² + 4a + 2) ] / (D(a))²Simplify the numerator inside the brackets:(a³ + 2a² + 2a + 4) - a(3a² + 4a + 2) = a³ + 2a² + 2a + 4 - 3a³ - 4a² - 2aCombine like terms:a³ - 3a³ = -2a³2a² - 4a² = -2a²2a - 2a = 0So, the numerator becomes:-2a³ - 2a² + 4Therefore, f'(a) = 36(-2a³ - 2a² + 4) / (D(a))²Set f'(a) = 0:36(-2a³ - 2a² + 4) = 0Since 36 ≠ 0, we have:-2a³ - 2a² + 4 = 0Divide both sides by -2:a³ + a² - 2 = 0So, we need to solve the cubic equation:a³ + a² - 2 = 0Let me try to find rational roots using the Rational Root Theorem. Possible rational roots are ±1, ±2.Test a=1: 1 + 1 - 2 = 0. Yes, a=1 is a root.So, we can factor (a - 1) out of the cubic:Using polynomial division or synthetic division:Divide a³ + a² - 2 by (a - 1):Coefficients: 1 | 1 | 0 | -2Wait, actually, the cubic is a³ + a² - 2, so coefficients are 1 (a³), 1 (a²), 0 (a), -2 (constant).Using synthetic division with root a=1:Bring down 1.Multiply by 1: 1.Add to next coefficient: 1 + 1 = 2.Multiply by 1: 2.Add to next coefficient: 0 + 2 = 2.Multiply by 1: 2.Add to last coefficient: -2 + 2 = 0.So, the cubic factors as (a - 1)(a² + 2a + 2) = 0Set each factor equal to zero:a - 1 = 0 => a = 1a² + 2a + 2 = 0 => discriminant = 4 - 8 = -4 < 0, so no real roots.Therefore, the only real root is a=1.So, critical point at a=1.Now, we need to check whether this critical point is a maximum.Since f(a) approaches 0 as a approaches 0 and as a approaches infinity, and there's only one critical point, it must be a maximum.Therefore, the maximum occurs at a=1.Now, let's find b and c:b = 6 / (a² + 2) = 6 / (1 + 2) = 6 / 3 = 2c = 6 / (a + 2) = 6 / (1 + 2) = 6 / 3 = 2So, a=1, b=2, c=2.Therefore, the maximum value of abc is 1*2*2=4.Wait, hold on. Earlier, when I thought about using AM-GM inequality, I thought the maximum might be 3, but with this method, I get abc=4.Let me double-check my calculations.First, when I found a=1, then b=6/(1+2)=2, c=6/(1+2)=2, so abc=1*2*2=4.But when I tried using AM-GM earlier, I thought the maximum was 3. Maybe I made a mistake there.Wait, let me go back to the AM-GM part.I had:(a² + 2)(a + 2) ≥ 4aBut when a=1, (1 + 2)(1 + 2)=9, which is equal to 4a=4*1=4? Wait, 9 is not equal to 4. So, my AM-GM step might have been incorrect.Wait, I think I misapplied AM-GM.Let me try that again.I had:a² + 2 ≥ 2√(2a²) = 2a√2And a + 2 ≥ 2√(2a)So, (a² + 2)(a + 2) ≥ (2a√2)(2√(2a)) = 4a√2 * √(2a) = 4a√(2a) * √2 = 4a * (2a)^{1/2} * 2^{1/2}Wait, this seems messy. Maybe it's not the best approach.Alternatively, perhaps I should not have used AM-GM and stick with the calculus approach.Given that the critical point at a=1 gives abc=4, and since it's the only critical point, it must be the maximum.Wait, but when I substituted a=2 earlier, I got abc=3, but that was under a different assumption. Maybe I confused myself.Wait, no, earlier when I thought a=2, b=1, c=3/2, giving abc=3, but that was under a different approach where I thought the maximum was 3, but actually, the correct approach via Lagrange multipliers gives a=1, b=2, c=2, abc=4.Therefore, the maximum value is 4.Wait, but let me verify with a=1:b(a² + 2) + c(a + 2) = 2*(1 + 2) + 2*(1 + 2) = 2*3 + 2*3 = 6 + 6 = 12, which satisfies the constraint.So, everything checks out.Therefore, the maximum value of abc is 4, achieved when a=1, b=2, c=2.I think my initial confusion came from misapplying the AM-GM inequality, but the calculus approach gives a clear answer.**Final Answer**The positive reals (a, b, c) that maximize (abc) are (boxed{1}), (boxed{2}), and (boxed{2}), respectively, giving the maximum value of (boxed{4}).**Final Answer**The maximum value of (abc) is (boxed{4}).

answer：Alright, so I have this problem about Maria walking down Swanson Avenue and passing four houses in a row, each painted a different color: green, blue, red, and pink. I need to figure out how many possible orderings of these houses there are based on the given conditions. Let me try to break this down step by step.First, let's list out the conditions:1. Maria passed the green house before the blue house.2. Maria passed the red house before the pink house.3. The green house was not next to the pink house.Okay, so we have four houses: green (G), blue (B), red (R), and pink (P). Each is a different color, and Maria passes them in some order. The conditions give us specific constraints on the order.Let me think about how to approach this. Since there are four houses, there are 4! = 24 possible orderings without any constraints. But with the given constraints, this number will be reduced.Let me consider the constraints one by one.**First Constraint: Green before Blue (G < B)**This means that in the ordering, wherever G is, B must come after it. So, for any arrangement, G cannot be after B.**Second Constraint: Red before Pink (R < P)**Similarly, R must come before P in the ordering.**Third Constraint: Green not next to Pink (G not adjacent to P)**This adds another layer of complexity. So, G and P cannot be next to each other in the sequence.Alright, so I need to find all possible permutations of G, B, R, P where:1. G comes before B.2. R comes before P.3. G and P are not adjacent.One way to approach this is to list all possible permutations that satisfy the first two constraints and then eliminate those that violate the third constraint.But listing all 24 permutations might be time-consuming. Maybe I can find a smarter way.Let me think about the positions of G and P first because they have a specific constraint between them.Since G cannot be next to P, their positions must be separated by at least one house.Given that there are four houses, let's consider the possible positions for G and P such that they are not adjacent.Let me denote the positions as 1, 2, 3, 4.Possible non-adjacent pairs for G and P:- G in position 1, P in position 3 or 4- G in position 2, P in position 4- G in position 3, P in position 1 (but since G must come before B, and if G is in position 3, B must be in position 4, which might conflict with P's position)- G in position 4, but then P cannot be in any position since G is last, but P must come after R, so this might not work.Wait, actually, since G must come before B, and R must come before P, G and R can be anywhere as long as their respective constraints are satisfied.But since G cannot be next to P, let's focus on their positions.Let me consider the possible positions for G and P:1. If G is in position 1, then P can be in positions 3 or 4.2. If G is in position 2, then P can be in position 4.3. If G is in position 3, then P can be in position 1, but since G must come before B, and if G is in position 3, B must be in position 4, which would mean P is in position 1, but P must come after R, so R would have to be in position 1 or 2, but P is in position 1, which would mean R is before P, which is okay, but let's see if that works.4. If G is in position 4, then P cannot be anywhere because P must come after R, but G is already last, so this might not be possible.Wait, actually, if G is in position 4, then B must be after G, but there are no positions after 4, so G cannot be in position 4. So G can only be in positions 1, 2, or 3.Similarly, P must come after R, so P cannot be in position 1.So, let's re-examine:1. G in position 1: - P can be in 3 or 4. - If P is in 3, then R must be in 2 (since R must come before P). - If P is in 4, then R can be in 2 or 3.2. G in position 2: - P can only be in 4. - Then R must be in 1 or 3.3. G in position 3: - P can be in 1, but as I thought earlier, if G is in 3, B must be in 4. - Then P is in 1, so R must be in 2 or 3, but G is in 3, so R must be in 2.Let me try to map this out.**Case 1: G in position 1**- P can be in 3 or 4.- **Subcase 1a: P in 3** - Then R must be in 2 (since R < P). - Then B must be in 4 (since G < B). - So the order is G, R, P, B.- **Subcase 1b: P in 4** - Then R can be in 2 or 3. - If R is in 2: - Then B must be in 3 or 4, but P is in 4, so B must be in 3. - So the order is G, R, B, P. - If R is in 3: - Then B must be in 4. - So the order is G, B, R, P.Wait, but in Subcase 1b, if R is in 3, then B must be in 4, which is okay, but we have to ensure that G is before B, which it is, and R is before P, which it is.So from Case 1, we have three possible orderings:1. G, R, P, B2. G, R, B, P3. G, B, R, PBut wait, in Subcase 1b, when R is in 3, the order is G, B, R, P. Is that valid? Let's check:- G before B: Yes, G is in 1, B in 2.- R before P: Yes, R in 3, P in 4.- G not next to P: G is in 1, P in 4, so they are separated by two houses. So yes, valid.So Case 1 gives us three orderings.**Case 2: G in position 2**- P can only be in 4.- Then R must be in 1 or 3.- **Subcase 2a: R in 1** - Then B must be in 3 or 4, but P is in 4, so B must be in 3. - So the order is R, G, B, P.- **Subcase 2b: R in 3** - Then B must be in 4. - So the order is R, G, B, P.Wait, that's the same as Subcase 2a. Wait, no, if R is in 3, then the order would be R, G, B, P, which is the same as Subcase 2a.Wait, no, if R is in 3, then the order is R, G, B, P, which is the same as Subcase 2a if R is in 1. Wait, no, if R is in 1, the order is R, G, B, P. If R is in 3, the order is G, R, B, P, but G is in 2, so it would be R, G, B, P.Wait, I'm getting confused. Let me clarify.If G is in position 2, and P is in 4, then R can be in 1 or 3.- If R is in 1: - Then the order is R, G, B, P.- If R is in 3: - Then the order is G, R, B, P.Wait, but G is in 2, so if R is in 3, the order would be R in 1, G in 2, R in 3? No, that doesn't make sense because R can't be in two places.Wait, no, if G is in 2, and R is in 3, then the order is:1: R2: G3: RBut R is only one house, so that's not possible. Wait, I think I made a mistake.If G is in 2, and R is in 3, then the order would be:1: ?2: G3: R4: PBut we have four houses: G, B, R, P.So if G is in 2, R is in 3, then position 1 must be B or something else.Wait, no, because R must come before P, and G must come before B.Wait, let's think again.If G is in 2, P is in 4, and R is in 3, then:- Position 1 must be B, because G is in 2, and B must come after G.But then the order would be:1: B2: G3: R4: PBut does this satisfy all constraints?- G before B: Yes, G is in 2, B in 1. Wait, no, G is in 2, B is in 1, which violates G before B.Ah, that's a problem. So if G is in 2, and R is in 3, then B must be after G, which would be in position 4, but P is already in 4. So that's not possible.Therefore, if G is in 2, and R is in 3, it's not possible because B would have to be in 4, but P is already in 4. So R cannot be in 3 in this case.Therefore, only Subcase 2a is valid:- R in 1, G in 2, B in 3, P in 4.So the order is R, G, B, P.So Case 2 gives us one ordering.**Case 3: G in position 3**- P can be in 1.- Then R must be in 2 (since R < P, and P is in 1, which is not possible because R must come before P, but P is in 1, which is the first position. So R cannot be in 1 because P is already in 1.Wait, that doesn't make sense. If P is in 1, then R must be in a position before 1, which is impossible because there are no positions before 1. Therefore, if G is in 3, P cannot be in 1 because R cannot come before P in that case.Alternatively, maybe I made a mistake earlier. Let me think again.If G is in 3, then P cannot be adjacent, so P cannot be in 2 or 4.Wait, P cannot be adjacent to G. If G is in 3, then P cannot be in 2 or 4.But P must come after R, so P must be in a position after R.If G is in 3, then P cannot be in 4 because that would make G and P adjacent (positions 3 and 4). So P cannot be in 4.Similarly, P cannot be in 2 because that would be adjacent to G in 3.Therefore, P cannot be in 2 or 4 if G is in 3, which leaves only position 1 for P.But as I thought earlier, if P is in 1, then R must be in a position before 1, which is impossible. Therefore, G cannot be in 3 because it would force P into an impossible position.Therefore, Case 3 is invalid.So, summarizing:- Case 1: G in 1 gives us three orderings.- Case 2: G in 2 gives us one ordering.- Case 3: G in 3 is invalid.Wait, but earlier in Case 1, I thought of three orderings, but let me check again.In Case 1:- G in 1, P in 3: G, R, P, B- G in 1, P in 4: - R in 2: G, R, B, P - R in 3: G, B, R, PSo that's three orderings.In Case 2:- G in 2, P in 4, R in 1: R, G, B, PSo that's one ordering.Therefore, total orderings are 3 + 1 = 4.Wait, but earlier I thought of three in Case 1 and one in Case 2, making four.But let me check if all these orderings satisfy all constraints.1. G, R, P, B: - G before B: Yes. - R before P: Yes. - G not next to P: G in 1, P in 3: separated by R, so yes.2. G, R, B, P: - G before B: Yes. - R before P: Yes. - G not next to P: G in 1, P in 4: separated by R and B, so yes.3. G, B, R, P: - G before B: Yes. - R before P: Yes. - G not next to P: G in 1, P in 4: separated by B and R, so yes.4. R, G, B, P: - G before B: Yes. - R before P: Yes. - G not next to P: G in 2, P in 4: separated by B, so yes.So all four orderings satisfy the constraints.Wait, but the answer choices only go up to 5, and the options are A)1, B)2, C)3, D)4, E)5.So according to my reasoning, the answer should be D)4.But wait, let me double-check because sometimes I might have missed something.Let me list all four orderings:1. G, R, P, B2. G, R, B, P3. G, B, R, P4. R, G, B, PAre there any other possible orderings?Let me think if there's another way to arrange them.What if G is in position 2, P in 4, and R in 3?Wait, that would be R in 3, G in 2, but then R is after G, which is okay, but R must come before P, which is in 4, so that's fine. But G is in 2, so B must be after G, which would be in 3 or 4. But P is in 4, so B would have to be in 3.So the order would be:1: ?2: G3: B4: PBut then R must be in 1 because R must come before P, but R is in 3 in this case, which is after G in 2, but R is in 3, which is before P in 4, so that's okay.Wait, but if G is in 2, R is in 3, then the order would be:1: ?2: G3: R4: PBut then position 1 must be B, because we have four houses: G, B, R, P.So the order would be:1: B2: G3: R4: PBut does this satisfy all constraints?- G before B: G is in 2, B is in 1. No, that's B before G, which violates the first constraint.Therefore, this ordering is invalid.So, that ordering doesn't work.Therefore, the only valid orderings are the four I found earlier.So, the answer should be D)4.But wait, the initial answer provided was B)2, which contradicts my reasoning.Let me see where I might have gone wrong.Wait, in the initial answer, the assistant considered only two orderings:- G-R-B-P- R-G-B-PBut according to my reasoning, there are four valid orderings.Let me check the initial answer's reasoning.The assistant considered:**Case 1: P is the 3rd house.**- G must be the 1st house.- The arrangement is G-R-B-P.**Case 2: P is the last house.**- G can be 1st or 2nd. - Subcase 2.1: G is 1st: G-R-B-P - Subcase 2.2: G is 2nd: R-G-B-PThus, the assistant concluded two orderings: G-R-B-P and R-G-B-P.But according to my reasoning, there are more orderings.Wait, perhaps the assistant missed some cases.Let me check the assistant's reasoning.The assistant considered:- If P is in 3, then G must be in 1, leading to G-R-B-P.- If P is in 4, then G can be in 1 or 2. - If G is in 1, then R and B can be arranged as R-B or B-R, but since R must come before P, and P is in 4, R can be in 2 or 3. - If R is in 2, then B is in 3: G-R-B-P. - If R is in 3, then B is in 2: G-B-R-P. - If G is in 2, then R must be in 1, leading to R-G-B-P.So, actually, the assistant should have considered G-B-R-P as another ordering.Therefore, the assistant missed that ordering.So, the correct number of orderings should be three: G-R-B-P, G-B-R-P, and R-G-B-P.Wait, but in my earlier reasoning, I found four orderings, including G-R-P-B.Wait, let me check if G-R-P-B is valid.- G before B: Yes, G in 1, B in 4.- R before P: Yes, R in 2, P in 3.- G not next to P: G in 1, P in 3: separated by R, so yes.So, G-R-P-B is valid.Therefore, the four orderings are:1. G-R-P-B2. G-R-B-P3. G-B-R-P4. R-G-B-PSo, the initial assistant's answer was incorrect because they missed two orderings.Therefore, the correct answer should be D)4.But wait, let me double-check if all four orderings are indeed valid.1. G-R-P-B: - G before B: Yes. - R before P: Yes. - G not next to P: Yes, separated by R.2. G-R-B-P: - G before B: Yes. - R before P: Yes. - G not next to P: Yes, separated by R and B.3. G-B-R-P: - G before B: Yes. - R before P: Yes. - G not next to P: Yes, separated by B and R.4. R-G-B-P: - G before B: Yes. - R before P: Yes. - G not next to P: Yes, separated by B.All four orderings satisfy the constraints.Therefore, the correct answer is D)4.

answer：First, I need to determine how many doughnuts were eaten by the staff members. There were 50 doughnuts served in total, and 12 were left uneaten. So, the number of doughnuts eaten is 50 minus 12, which equals 38.Next, I know that each staff member ate 2 doughnuts. To find out how many staff members attended the meeting, I can divide the total number of doughnuts eaten by the number of doughnuts each staff member consumed. That is, 38 doughnuts eaten divided by 2 doughnuts per staff member equals 19 staff members.Therefore, there were 19 staff members at the meeting.

answer：First, identify the total number of pens Morgan has, which is 168.Next, note the number of blue pens (45) and black pens (58).Add the number of blue and black pens together: 45 + 58 = 103.Subtract the sum of blue and black pens from the total number of pens to find the number of red pens: 168 - 103 = 65.Therefore, Morgan has 65 red pens.