💡First, I need to consider the constraints given in the problem. Student A cannot be at either end of the row, and students B and C must be adjacent.To handle the adjacency requirement for B and C, I can treat them as a single unit. This reduces the problem from arranging six individual students to arranging five units (the BC unit plus the other four students).Next, I need to account for the restriction on student A. Since A cannot be at either end, there are three possible positions for A within the row.Finally, I calculate the number of ways to arrange the remaining students. With A placed in one of the three middle positions, the remaining four units can be arranged in 4! ways.By multiplying the number of ways to arrange B and C (2), the number of positions for A (3), and the number of ways to arrange the remaining students (24), I find the total number of valid arrangements.

💡To find the coefficient of ( x^{3} ) in the expansion of ( left(2 - xright)^{5} ), we can use the binomial theorem. The binomial theorem states that:[(a + b)^{n} = sum_{k=0}^{n} {C}_{n}^{k} a^{n - k} b^{k}]In this case, ( a = 2 ), ( b = -x ), and ( n = 5 ). We are interested in the term where the exponent of ( x ) is 3, which corresponds to ( k = 3 ).1. **Calculate the binomial coefficient**: [ {C}_{5}^{3} = frac{5!}{3!(5 - 3)!} = frac{120}{6 times 2} = 10 ]2. **Compute the powers of ( a ) and ( b )**: [ a^{5 - 3} = 2^{2} = 4 ] [ b^{3} = (-x)^{3} = -x^{3} ]3. **Multiply the results together**: [ {C}_{5}^{3} times 2^{2} times (-x)^{3} = 10 times 4 times (-x^{3}) = 40 times (-x^{3}) = -40x^{3} ]Therefore, the coefficient of ( x^{3} ) in the expansion is ( -40 ).

💡First, identify the relationships between the coins based on the given conditions:- ( C ) covers both ( A ) and ( D ), indicating ( C ) is above ( A ) and ( D ).- ( A ) partially covers ( B ), meaning ( A ) is above ( B ).- ( D ) covers ( B ) but not ( E ), so ( D ) is above ( B ) but below ( E ).- ( E ) is covered by ( C ) and does not cover any other coin, placing ( E ) below ( C ) but above ( A ) and ( D ).From these relationships, we can establish the following order:1. ( C ) is at the top since it covers ( A ), ( D ), and ( E ).2. ( E ) is next, as it is covered by ( C ) but does not cover any other coin.3. ( A ) follows ( E ), as ( A ) covers ( B ).4. ( D ) is below ( A ), as ( D ) covers ( B ) but is covered by ( C ).5. ( B ) is at the bottom, being covered by both ( A ) and ( D ).Thus, the order from top to bottom is ( C, E, A, D, B ).

💡First, I need to determine the area of the floor by multiplying the length and width of the room.Given:- Length = 5 meters- Width = 4.75 metersCalculating the area:Area = Length × Width = 5 m × 4.75 m = 23.75 square metersNext, I'll calculate the total cost by multiplying the area by the cost per square meter.Given:- Cost per square meter = ₹900Calculating the total cost:Total Cost = Area × Cost per square meter = 23.75 sq.m × ₹900/sq.m = ₹21,375Therefore, the total cost to pave the floor is ₹21,375.

💡Okay, so I have this problem that says I need to prove that the sum of the cubes of all numbers from 1 to 1001 is equal to zero. Hmm, that's interesting. At first glance, it seems a bit odd because when I think of summing cubes, I don't immediately think of getting zero. But maybe there's something special about the number 1001 that makes this happen.Let me start by writing out what the sum actually is. The sum S would be:S = 1³ + 2³ + 3³ + ... + 500³ + 501³ + ... + 999³ + 1000³ + 1001³Wait, but 1001 is included here. So, does that mean I'm summing from 1 all the way up to 1001? That's 1001 terms. But I'm supposed to show that this sum equals zero. That's surprising because all these cubes are positive numbers, right? So how can their sum be zero?Maybe there's something about modular arithmetic or properties of numbers modulo 1001 that I need to consider. I remember that in modular arithmetic, numbers wrap around after a certain modulus. So, for example, in modulo 1001, 1001 is equivalent to 0. That means 1000 is equivalent to -1, 999 is equivalent to -2, and so on.So, if I think of each number from 1 to 1000 in terms of modulo 1001, they can be paired with their negative counterparts. For instance, 1 is equivalent to -1000, 2 is equivalent to -999, and so on, up to 500 being equivalent to -501. This pairing might be useful because when I cube these numbers, their cubes might cancel each other out.Let me test this idea with a few examples. If I take 1³ and 1000³, since 1000 is equivalent to -1 modulo 1001, then 1000³ is equivalent to (-1)³, which is -1. So, 1³ + 1000³ would be 1 + (-1) = 0. That's promising!Similarly, let's take 2³ and 999³. 999 is equivalent to -2 modulo 1001, so 999³ is (-2)³ = -8. Therefore, 2³ + 999³ = 8 + (-8) = 0. Again, that cancels out.This pattern seems to hold for each pair. If I pair each number k with 1001 - k, then k³ + (1001 - k)³ should equal zero because (1001 - k) is equivalent to -k modulo 1001, and (-k)³ = -k³.So, if I pair all the numbers from 1 to 1000 like this, each pair sums to zero. That leaves me with the number 1001³. But since 1001 is equivalent to 0 modulo 1001, 1001³ is just 0³ = 0.Therefore, the entire sum S would be the sum of all these pairs, each of which is zero, plus 0 from 1001³. So, S = 0 + 0 + ... + 0 = 0.Wait, but I should double-check this. Let me think about how many pairs there are. From 1 to 1000, there are 1000 numbers, which makes 500 pairs. Each pair sums to zero, so 500 * 0 = 0. Adding 1001³, which is 0, still gives me 0. That seems correct.But just to be thorough, let me consider the general case. Suppose I have a modulus n, and I want to sum the cubes of all numbers from 1 to n. If n is odd, like 1001, then the numbers can be perfectly paired as k and n - k, where k ranges from 1 to (n-1)/2. Each pair would sum to zero because (n - k) ≡ -k mod n, and thus (n - k)³ ≡ (-k)³ = -k³ mod n. Therefore, the sum of the cubes from 1 to n would be zero modulo n.In this case, n is 1001, which is odd, so this property applies. Hence, the sum of the cubes from 1 to 1001 is indeed zero.I think that makes sense. By pairing each number with its negative counterpart modulo 1001, their cubes cancel each other out, leading to the total sum being zero. It's a neat application of modular arithmetic properties.

💡First, I need to determine the length of the arc corresponding to a 120° angle in a circular curve with a radius of 10 meters.I recall that the formula for the length of an arc is given by:[ l = r times alpha ]where ( l ) is the arc length, ( r ) is the radius, and ( alpha ) is the central angle in radians.Since the given angle is in degrees, I need to convert it to radians. There are ( 2pi ) radians in 360°, so:[ 120° = frac{120}{360} times 2pi = frac{1}{3} times 2pi = frac{2pi}{3} text{ radians} ]Now, I can plug the values into the arc length formula:[ l = 10 times frac{2pi}{3} = frac{20pi}{3} text{ meters} ]Therefore, the length of the curve corresponding to a 120° angle is (frac{20pi}{3}) meters.