💡Okay, so I have this function f(x) that's defined in a piecewise manner. It says that if x is greater than 0, then f(x) is log base 2 of x. But if x is less than or equal to 0, then f(x) is equal to f(x + 3). I need to find the value of f(-10). Hmm, that seems a bit tricky at first glance, but let me break it down step by step.Alright, starting with f(-10). Since -10 is less than or equal to 0, according to the function definition, I should use the second case, which is f(x + 3). So, f(-10) becomes f(-10 + 3), which is f(-7). Okay, so now I need to find f(-7).Again, -7 is still less than or equal to 0, so I apply the same rule. f(-7) becomes f(-7 + 3), which is f(-4). Moving on, f(-4) is still in the negative range, so I add 3 again. f(-4) becomes f(-4 + 3) = f(-1). Now, f(-1) is still less than or equal to 0, so I add 3 once more. f(-1) becomes f(-1 + 3) = f(2).Alright, now I have f(2). This time, 2 is greater than 0, so I use the first case of the function, which is log base 2 of x. So, f(2) is log base 2 of 2. I remember that log base a of a is always 1 because a to the power of 1 is a. So, log base 2 of 2 is 1.Wait, let me make sure I didn't skip any steps or make a mistake. Starting from f(-10), I added 3 each time until I got to a positive number. Each step was f(-10) = f(-7), then f(-7) = f(-4), then f(-4) = f(-1), and finally f(-1) = f(2). That seems correct. And since 2 is positive, I used the log function, which gave me 1.Is there another way to think about this? Maybe using recursion or some kind of pattern? Let's see. If I keep adding 3 to x until it becomes positive, how many times do I add 3? Starting from -10, adding 3 four times gets me to 2. So, f(-10) = f(-7) = f(-4) = f(-1) = f(2). Yeah, that's consistent with what I did before.I wonder if there's a formula or a shortcut for this kind of recursive function. If x is negative, I keep adding 3 until it's positive, and then take the log base 2. So, in general, for any negative x, f(x) = log base 2 of (x + 3n), where n is the smallest integer such that x + 3n > 0.Let me test this with x = -10. I need to find n such that -10 + 3n > 0. Solving for n: 3n > 10, so n > 10/3, which is approximately 3.333. Since n has to be an integer, the smallest n is 4. So, f(-10) = log base 2 of (-10 + 3*4) = log base 2 of 2, which is 1. That matches my earlier result.Is there a way to generalize this for any x? If x is negative, f(x) = log base 2 of (x + 3n), where n is the smallest integer such that x + 3n > 0. So, n = ceiling of (|x| / 3). For x = -10, |x| = 10, 10 / 3 ≈ 3.333, ceiling of that is 4. So, n = 4, and f(-10) = log base 2 of (-10 + 12) = log base 2 of 2 = 1.Wait, let me make sure I didn't make a mistake in the generalization. If x is negative, say x = -k where k > 0, then we need to find n such that -k + 3n > 0, which simplifies to 3n > k, so n > k/3. Since n must be an integer, n is the smallest integer greater than k/3, which is the ceiling of k/3. So, f(x) = log base 2 of (x + 3n), where n = ceiling(k/3) = ceiling(|x|/3).Applying this to x = -10, k = 10, ceiling(10/3) = 4, so f(-10) = log base 2 of (-10 + 12) = log base 2 of 2 = 1. Yep, that still works.I think I've got it. The function essentially shifts the input x by multiples of 3 until it becomes positive, and then takes the log base 2 of that positive number. So, for f(-10), shifting by 4 times 3 gets us to 2, and log base 2 of 2 is 1.Just to double-check, let's try another example. Suppose x = -4. According to the function, f(-4) = f(-1) = f(2) = log base 2 of 2 = 1. Using the formula, n = ceiling(4/3) = 2, so f(-4) = log base 2 of (-4 + 6) = log base 2 of 2 = 1. It matches.Another example: x = -1. f(-1) = f(2) = 1. Using the formula, n = ceiling(1/3) = 1, so f(-1) = log base 2 of (-1 + 3) = log base 2 of 2 = 1. Correct.What about x = 0? f(0) = f(3) = log base 2 of 3. Using the formula, n = ceiling(0/3) = 0, so f(0) = log base 2 of (0 + 0) = log base 2 of 0. Wait, that's undefined. Hmm, but according to the function definition, f(0) = f(3), and f(3) = log base 2 of 3, which is defined. So, maybe my formula needs adjustment for x = 0. Since x = 0, n = ceiling(0/3) = 0, but x + 3n = 0, which is not positive. So, perhaps n should be ceiling((x)/3) where x is negative. Wait, for x = 0, it's already non-negative, so f(0) = f(3). So, maybe the formula is only for x < 0.In any case, for x = -10, the process works, and the answer is 1.

💡Okay, so I have this geometry problem here about a triangle ABC with medians BK and CN intersecting at point M. I need to figure out the maximum number of sides of quadrilateral ANMK that can have a length of 1. Hmm, let me start by drawing a rough sketch of triangle ABC with the medians BK and CN. First, I remember that in a triangle, the medians intersect at the centroid, which divides each median into a ratio of 2:1. So, point M is the centroid here, meaning BM is twice as long as MK, and CM is twice as long as MN. That might be useful later.Now, quadrilateral ANMK has four sides: AN, NM, MK, and KA. I need to see how many of these can be equal to 1. Let me consider each side one by one.Starting with AN: If AN is 1, then since N is the midpoint of AB, that would mean AB is 2. Similarly, if NM is 1, and since M divides CN in a 2:1 ratio, that would mean CM is 2 and MN is 1. Wait, but if NM is 1, then CN is 3 because CM is twice MN. Hmm, that might complicate things.What about MK? If MK is 1, then since M divides BK in a 2:1 ratio, BK would be 3. Similarly, if KA is 1, then since K is the midpoint of AC, AC would be 2. Let me try to see if I can have three sides of ANMK equal to 1. Suppose AN, NM, and MK are all 1. Then, as I thought earlier, AB would be 2, CN would be 3, BK would be 3, and AC would be 2. But wait, if AB is 2 and AC is 2, then triangle ABC would be isosceles with AB = AC. But if that's the case, then the medians BK and CN would also have some equal properties. But hold on, if AB = AC, then the centroid M would lie along the altitude from A to BC. However, I'm not sure if that necessarily causes a problem. Maybe I need to check the lengths more carefully. If AN = 1, then since N is the midpoint, AB = 2. If NM = 1, then since M is the centroid, CN = 3. Similarly, if MK = 1, then BK = 3. But then, in triangle ABC, sides AB = 2, AC = 2 (since KA = 1, so AC = 2), and BC? Hmm, I don't know BC yet. Let me see if I can find BC using the medians. The formula for the length of a median is m_a = (1/2)√(2b² + 2c² - a²). So, for median BK, which is 3, it would be (1/2)√(2AB² + 2BC² - AC²). Plugging in AB = 2, AC = 2, and BK = 3, we get:3 = (1/2)√(2*(2)² + 2*BC² - (2)²)3 = (1/2)√(8 + 2BC² - 4)3 = (1/2)√(4 + 2BC²)6 = √(4 + 2BC²)36 = 4 + 2BC²32 = 2BC²16 = BC²BC = 4So BC is 4. Now, let's check if this works with the other median CN, which is also 3. Using the median formula again for CN:3 = (1/2)√(2AB² + 2AC² - BC²)3 = (1/2)√(2*(2)² + 2*(2)² - (4)²)3 = (1/2)√(8 + 8 - 16)3 = (1/2)√(0)3 = 0Wait, that doesn't make sense. I must have made a mistake somewhere. Maybe assuming three sides equal to 1 is impossible because it leads to a contradiction. Let me try another approach. Suppose only two sides of quadrilateral ANMK are equal to 1. Let's say AN = 1 and MK = 1. Then, AB = 2, BK = 3, and AC = 2. Then, as before, BC would be 4. Now, let's check the median CN. Using the median formula again for CN:CN = (1/2)√(2AB² + 2AC² - BC²)CN = (1/2)√(2*(2)² + 2*(2)² - (4)²)CN = (1/2)√(8 + 8 - 16)CN = (1/2)√(0)CN = 0Again, that's not possible. Hmm, maybe I need to consider different sides being equal to 1. What if AN = 1 and NM = 1? Then, AB = 2, CN = 3, and since M is the centroid, CM = 2 and MN = 1. Then, using the median formula for BK:BK = (1/2)√(2AB² + 2BC² - AC²)But I don't know AC yet. Wait, if AN = 1 and N is the midpoint, then AB = 2. If NM = 1, then CN = 3, so AC can be found using the median formula for CN:CN = (1/2)√(2AB² + 2AC² - BC²)3 = (1/2)√(8 + 2AC² - BC²)6 = √(8 + 2AC² - BC²)36 = 8 + 2AC² - BC²28 = 2AC² - BC²But I also know that BK is a median, so:BK = (1/2)√(2AB² + 2BC² - AC²)But I don't know BK yet. Wait, if MK = 1, then BK = 3, so:3 = (1/2)√(8 + 2BC² - AC²)6 = √(8 + 2BC² - AC²)36 = 8 + 2BC² - AC²28 = 2BC² - AC²Now I have two equations:28 = 2AC² - BC²28 = 2BC² - AC²Let me solve these simultaneously. From the first equation:2AC² - BC² = 28From the second equation:2BC² - AC² = 28Let me rearrange the first equation:2AC² = BC² + 28AC² = (BC² + 28)/2Now plug this into the second equation:2BC² - (BC² + 28)/2 = 28Multiply both sides by 2 to eliminate the denominator:4BC² - (BC² + 28) = 564BC² - BC² - 28 = 563BC² = 84BC² = 28BC = √28 = 2√7Now, plug BC² = 28 into the first equation:2AC² - 28 = 282AC² = 56AC² = 28AC = √28 = 2√7So, AC = 2√7 and BC = 2√7. That means triangle ABC is isosceles with AB = 2 and AC = BC = 2√7. Let me check if this works with the medians.Median CN is 3, which we already used. Median BK is 3 as well. So, both medians are 3. Let me verify the lengths:AN = 1, NM = 1, MK = 1? Wait, no, we only assumed AN = 1 and NM = 1, but MK would be 1 as well if BK = 3. But earlier, when I tried three sides, it led to a contradiction. Maybe in this specific case, it's possible?Wait, no, because if AC = 2√7 and AB = 2, then the median BK should be:BK = (1/2)√(2AB² + 2BC² - AC²)= (1/2)√(2*(2)² + 2*(2√7)² - (2√7)²)= (1/2)√(8 + 2*28 - 28)= (1/2)√(8 + 56 - 28)= (1/2)√(36)= (1/2)*6= 3Okay, that works. So, BK is indeed 3. Similarly, CN is 3 as well. So, in this case, quadrilateral ANMK has sides AN = 1, NM = 1, MK = 1, and KA. Wait, KA is half of AC, which is √28 / 2 = √7 ≈ 2.645, which is not 1. So, only three sides are 1: AN, NM, and MK. But earlier, when I tried this, it led to a contradiction with the median formula. But now, it seems to work because AC and BC are both 2√7, making the medians equal to 3.Wait, but earlier, when I assumed AN = 1, NM = 1, and MK = 1, I ended up with CN = 0, which was a problem. But now, with AC = BC = 2√7, it seems to hold. Maybe my earlier mistake was not considering that AC and BC could be longer, allowing the medians to be 3 without contradiction.But hold on, if AN = 1, NM = 1, and MK = 1, then quadrilateral ANMK has three sides equal to 1. Is that possible? Let me visualize it. Point A, then N is the midpoint of AB, so AN = 1, then M is the centroid, so NM = 1, and K is the midpoint of AC, so MK = 1. Hmm, but in reality, the distances might not all be 1 because the positions of the points are interdependent.Wait, maybe I'm overcomplicating this. Let me try to assign coordinates to the triangle to see if it's possible.Let me place point A at (0, 0). Let me assume AB is along the x-axis, so point B is at (2, 0) since AN = 1 and N is the midpoint. Now, point C is somewhere in the plane. Since AC = 2√7, point C must be at (x, y) such that the distance from A to C is 2√7, so x² + y² = (2√7)² = 28.Also, since BC = 2√7, the distance from B (2,0) to C (x,y) is also 2√7, so (x - 2)² + y² = 28.Subtracting the two equations:(x - 2)² + y² - (x² + y²) = 28 - 28x² - 4x + 4 + y² - x² - y² = 0-4x + 4 = 0-4x = -4x = 1So, x = 1. Plugging back into x² + y² = 28:1 + y² = 28y² = 27y = ±3√3So, point C is at (1, 3√3) or (1, -3√3). Let's take (1, 3√3) for simplicity.Now, let's find the coordinates of M, the centroid. The centroid is the average of the coordinates of A, B, and C.So, M_x = (0 + 2 + 1)/3 = 3/3 = 1M_y = (0 + 0 + 3√3)/3 = √3So, M is at (1, √3).Now, let's find the coordinates of K, the midpoint of AC. Since A is at (0,0) and C is at (1, 3√3), K is at ((0 + 1)/2, (0 + 3√3)/2) = (0.5, (3√3)/2).Now, let's compute the lengths of the sides of quadrilateral ANMK.AN: From A(0,0) to N(1,0). Distance is √[(1 - 0)² + (0 - 0)²] = √1 = 1. Good.NM: From N(1,0) to M(1, √3). Distance is √[(1 - 1)² + (√3 - 0)²] = √[0 + 3] = √3 ≈ 1.732. Not 1.Wait, that's not 1. Hmm, so NM is not 1. But I thought earlier that if AN = 1 and NM = 1, it would work. But according to these coordinates, NM is √3, not 1. So, my earlier assumption was wrong.Wait, maybe I made a mistake in assigning coordinates. Let me check.I set AB = 2, so A(0,0), B(2,0). Then, since AC = 2√7, point C is at (1, 3√3). Then, centroid M is at (1, √3). Midpoint N is at (1,0), and midpoint K is at (0.5, (3√3)/2).So, NM is from (1,0) to (1, √3), which is indeed √3. So, NM is not 1. Therefore, in this configuration, only AN = 1, and MK is from M(1, √3) to K(0.5, (3√3)/2). Let's compute MK:MK_x = 0.5 - 1 = -0.5MK_y = (3√3)/2 - √3 = (3√3)/2 - (2√3)/2 = (√3)/2Distance MK = √[(-0.5)² + (√3/2)²] = √[0.25 + 0.75] = √1 = 1. Okay, so MK is 1.So, in this configuration, AN = 1 and MK = 1. What about KA? From K(0.5, (3√3)/2) to A(0,0):KA_x = 0.5 - 0 = 0.5KA_y = (3√3)/2 - 0 = (3√3)/2Distance KA = √[(0.5)² + ( (3√3)/2 )²] = √[0.25 + (27/4)] = √[0.25 + 6.75] = √7 ≈ 2.645. Not 1.So, in this case, quadrilateral ANMK has two sides equal to 1: AN and MK. Wait, but earlier I thought NM could be 1, but in this configuration, it's √3. Maybe if I adjust the triangle differently, I can get NM to be 1 as well. Let me try.Suppose I want NM = 1. Since N is at (1,0) and M is at (1, √3), the distance is √3. To make NM = 1, I need M to be closer to N. But M is the centroid, so it's determined by the coordinates of A, B, and C. If I move point C closer to the x-axis, maybe M will move down, making NM shorter.Let me try placing point C at (1, h), where h is less than 3√3. Then, centroid M would be at (1, h/3). Then, NM would be the distance from (1,0) to (1, h/3), which is h/3. If I set h/3 = 1, then h = 3. So, point C would be at (1, 3). Let's see if this works.So, point C is at (1,3). Then, AC is the distance from (0,0) to (1,3):AC = √[(1)^2 + (3)^2] = √(1 + 9) = √10 ≈ 3.162. So, KA, being half of AC, would be √10 / 2 ≈ 1.581, not 1.But let's check the medians. Median CN is from C(1,3) to N(1,0). That's just a vertical line of length 3. So, CN = 3. Then, since M is the centroid, it divides CN into CM = 2 and MN = 1, which is what we wanted.Now, let's compute the length of median BK. Point B is at (2,0), and K is the midpoint of AC, which is at ((0 + 1)/2, (0 + 3)/2) = (0.5, 1.5). So, median BK is from B(2,0) to K(0.5,1.5). Let's compute its length:BK_x = 0.5 - 2 = -1.5BK_y = 1.5 - 0 = 1.5Distance BK = √[(-1.5)^2 + (1.5)^2] = √[2.25 + 2.25] = √4.5 ≈ 2.121. So, BK is not 3, which contradicts our earlier assumption that BK = 3 because MK = 1 implies BK = 3.Wait, so if I set C at (1,3) to make NM = 1, then BK becomes approximately 2.121, not 3. Therefore, MK, which is half of BK, would be approximately 1.06, not exactly 1. So, in this case, MK is not exactly 1, but close. Hmm, maybe I can adjust h slightly to make BK = 3.Let me set point C at (1, h). Then, K is at (0.5, h/2). Then, median BK is from B(2,0) to K(0.5, h/2). The length of BK is:BK = √[(0.5 - 2)^2 + (h/2 - 0)^2] = √[(-1.5)^2 + (h/2)^2] = √[2.25 + (h²)/4]We want BK = 3, so:√[2.25 + (h²)/4] = 3Square both sides:2.25 + (h²)/4 = 9(h²)/4 = 6.75h² = 27h = 3√3 ≈ 5.196Wait, that's the same as before. So, if I set h = 3√3, then BK = 3, which makes MK = 1. But in that case, NM is √3 ≈ 1.732, not 1. So, I can't have both NM = 1 and MK = 1 unless I adjust h, but adjusting h to make NM = 1 messes up BK.Therefore, it seems that I can have at most two sides of quadrilateral ANMK equal to 1: AN and MK. Because if I try to make NM = 1, it either messes up the length of BK or causes a contradiction in the triangle's properties.Wait, but earlier when I set C at (1,3√3), I had AN = 1, MK = 1, and NM = √3. So, only two sides are 1. If I try to set another side, say KA = 1, then AC would be 2, but that leads to a contradiction in the median lengths.Alternatively, if I set KA = 1, then AC = 2, and K is at (0.5, something). Then, median BK would be from B(2,0) to K(0.5, something). Let's compute the length of BK in that case.If AC = 2, then point C is somewhere such that AC = 2. Let me place C at (x,y) with x² + y² = 4. Then, midpoint K is at (x/2, y/2). Then, median BK is from B(2,0) to K(x/2, y/2). The length of BK is:BK = √[(x/2 - 2)^2 + (y/2 - 0)^2] = √[(x/2 - 2)^2 + (y/2)^2]We want BK = 3, so:√[(x/2 - 2)^2 + (y/2)^2] = 3Square both sides:(x/2 - 2)^2 + (y/2)^2 = 9Expand:(x²/4 - 2x + 4) + y²/4 = 9Multiply through by 4:x² - 8x + 16 + y² = 36But x² + y² = 4, so substitute:4 - 8x + 16 = 36-8x + 20 = 36-8x = 16x = -2So, x = -2. Then, from x² + y² = 4:(-2)^2 + y² = 44 + y² = 4y² = 0y = 0So, point C is at (-2, 0). But then, triangle ABC has points A(0,0), B(2,0), and C(-2,0). That's a degenerate triangle with all points on the x-axis. So, that's not possible because the medians would collapse onto the x-axis, and the centroid would be at (0,0), which is point A. That doesn't make sense because M would coincide with A, making quadrilateral ANMK degenerate.Therefore, setting KA = 1 leads to a degenerate triangle, which is not allowed. So, KA cannot be 1.So, in conclusion, the maximum number of sides of quadrilateral ANMK that can have a length of 1 is two: AN and MK. I can't have three sides equal to 1 without causing contradictions or degenerate triangles.

💡First, I recognize that a nonagon has 9 vertices. To find the number of distinct diagonals, I can use the formula for the number of diagonals in a convex polygon, which is (frac{n(n-3)}{2}), where (n) is the number of sides.Substituting (n = 9) into the formula:[frac{9(9 - 3)}{2} = frac{9 times 6}{2} = frac{54}{2} = 27]Therefore, there are 27 distinct diagonals in a convex nonagon.

💡Okay, so I have this problem where there are 8 books on a shelf, and I need to figure out how many ways I can select 5 books from them. But there's a catch: the selection must always include a specific book, let's call it Book X. The order in which I select the books doesn't matter, so it's a combination problem, not a permutation one.Alright, let's start by understanding what combinations are. Combinations are used when the order doesn't matter. For example, if I'm choosing books to take on a trip, it doesn't matter what order I pack them in; what matters is which books I end up with. So, the formula for combinations is:[binom{n}{k} = frac{n!}{k!(n - k)!}]where ( n ) is the total number of items, and ( k ) is the number of items we want to choose.In this problem, ( n = 8 ) because there are 8 books on the shelf, and ( k = 5 ) because we need to select 5 books. But wait, there's a condition: Book X must always be included in the selection. That changes things a bit.If Book X has to be included, then effectively, we're not choosing 5 books out of 8; instead, we're choosing 4 books out of the remaining 7. Because one of the 5 books is already fixed as Book X, we just need to select the other 4 from the remaining 7 books.So, now the problem simplifies to finding the number of ways to choose 4 books from 7. Using the combination formula again, we have:[binom{7}{4} = frac{7!}{4!(7 - 4)!} = frac{7!}{4! cdot 3!}]Calculating the factorials:- ( 7! = 7 times 6 times 5 times 4 times 3 times 2 times 1 = 5040 )- ( 4! = 4 times 3 times 2 times 1 = 24 )- ( 3! = 3 times 2 times 1 = 6 )Plugging these back into the formula:[binom{7}{4} = frac{5040}{24 times 6} = frac{5040}{144} = 35]So, there are 35 ways to select 5 books from the shelf that include Book X.Wait a second, let me double-check that. If I have 8 books and I need to choose 5, with one being fixed, then the number of combinations should be equal to choosing 4 from the remaining 7. That makes sense because one spot is already taken by Book X.Let me think about it another way. If I didn't have the condition of including Book X, the total number of ways to choose 5 books from 8 would be:[binom{8}{5} = frac{8!}{5! cdot 3!} = frac{40320}{120 times 6} = frac{40320}{720} = 56]So, without any restrictions, there are 56 ways. But since we have to include Book X, it's fewer. Specifically, it's the total number of combinations without restrictions minus the number of combinations that exclude Book X.The number of combinations that exclude Book X is the same as choosing all 5 books from the remaining 7 books:[binom{7}{5} = frac{7!}{5! cdot 2!} = frac{5040}{120 times 2} = frac{5040}{240} = 21]So, the number of valid combinations that include Book X is:[56 - 21 = 35]That matches my earlier calculation. So, it seems consistent.Another way to think about it is that for each combination that includes Book X, we're essentially pairing Book X with every possible group of 4 books from the remaining 7. Since the order doesn't matter, each group of 4 books combined with Book X gives a unique combination.Let me try to visualize this. Imagine the shelf with 8 books, and Book X is one of them. If I fix Book X as part of my selection, I need to pick 4 more books from the other 7. Each of these selections is unique because the combination formula accounts for all possible unique groupings without considering order.I can also think about smaller numbers to test this logic. Suppose there are 3 books on the shelf, and I need to choose 2, with one specific book always included. Using the same logic, I would fix that specific book and choose 1 more from the remaining 2. The number of ways would be:[binom{2}{1} = 2]Indeed, if the books are A, B, and C, and I must include A, the possible selections are AB and AC, which are 2 ways. This matches the calculation.Similarly, if I have 4 books and need to choose 3 with one specific book included, the number of ways would be:[binom{3}{2} = 3]Which makes sense: if the books are A, B, C, D, and I must include A, the selections are ABC, ABD, and ACD.This reinforces that the method of fixing one book and choosing the rest from the remaining is correct.Therefore, returning to the original problem, with 8 books and needing to choose 5 with Book X included, the number of ways is indeed:[binom{7}{4} = 35]I feel confident about this answer now, having checked it through different angles and smaller examples.

💡First, I need to solve the equation (2x - 1 = 1). To isolate the term with (x), I'll start by adding 1 to both sides of the equation. This will eliminate the constant term on the left side.After adding 1 to both sides, the equation becomes (2x = 2). Now, to solve for (x), I need to get rid of the coefficient 2 that's multiplying (x). I'll do this by dividing both sides of the equation by 2.Finally, dividing both sides by 2 gives me (x = 1). This is the solution to the equation.

💡First, treat products A and B as a single unit since they must be together. This gives us four units to arrange: AB, C, D, and E.The number of ways to arrange these four units is (4!).Since A and B can be in two different orders within their unit, we multiply by 2, resulting in (2 times 4! = 48) arrangements.Next, we need to subtract the arrangements where C and D are next to each other. Treat C and D as a single unit, giving us three units: AB, CD, and E.The number of ways to arrange these three units is (3!).Considering the internal arrangements of A and B and C and D, we multiply by (2 times 2), resulting in (2 times 2 times 3! = 24) arrangements to subtract.Finally, subtract the unwanted arrangements from the total: (48 - 24 = 24).Thus, the number of acceptable arrangements is 24.