Seven boys and two girls are to sit together on a bench. In how many ways can they arrange themselves so that the girls do not sit next to each other?

Here are the solutions to the mathematics questions. 2. Seven boys and two girls are to sit together on a bench. In how many ways can they arrange themselves so that the girls do not sit next to each other? Step 1: Arrange the 7 boys. The 7 boys can be arranged in 7! ways. 7! = 7 × 6 × 5 × 4 × 3 × 2 × 1 = 5040 Step 2: Create gaps for the girls. When the 7 boys are arranged, there are 8 possible positions (gaps) where the girls can sit so that they are not next to each other. Let B represent a boy and _ represent a gap: \_ B \_ B \_ B \_ B \_ B \_ B \_ B \_ Step 3: Place the 2 girls in 2 of these 8 gaps. The number of ways to choose 2 distinct gaps out of 8 and arrange the 2 distinct girls in them is given by the permutation formula P(n, k) = (n!)/((n-k)!). P(8, 2) = (8!)/((8-2)!) = (8!)/(6!) = 8 × 7 = 56 Step 4: Calculate the total number of arrangements. Multiply the number of ways to arrange the boys by the number of ways to place the girls. Total ways = 7! × P(8, 2) Total ways = 5040 × 56 Total ways = 282240 The girls can arrange themselves in 282240 ways so that they do not sit next to each other. 3. Express (5x^2 + 2x + 1)/((3x+1)(1+x^2)) in partial fractions. Step 1: Set up the partial fraction decomposition. The denominator has a linear factor (3x+1) and an irreducible quadratic factor (1+x^2). (5x^2 + 2x + 1)/((3x+1)(1+x^2)) = (A)/(3x+1) + (Bx+C)/(1+x^2) Step 2: Clear the denominators. Multiply both sides by (3x+1)(1+x^2): 5x^2 + 2x + 1 = A(1+x^2) + (Bx+C)(3x+1) Step 3: Solve for A, B, and C. To find A, let 3x+1 = 0, which means x = -(1)/(3). Substitute this value into the equation: 5(-(1)/(3))^2 + 2(-(1)/(3)) + 1 = A(1+(-(1)/(3))^2) + (B(-(1)/(3))+C)(0) 5((1)/(9)) - (2)/(3) + 1 = A(1+(1)/(9)) (5)/(9) - (6)/(9) + (9)/(9) = A((10)/(9)) (8)/(9) = A((10)/(9)) A = (8)/(9) × (9)/(10) = (8)/(10) = (4)/(5) So, A = (4)/(5). Now, expand the right side of the equation and equate coefficients: 5x^2 + 2x + 1 = A + Ax^2 + 3Bx^2 + Bx + 3Cx + C 5x^2 + 2x + 1 = (A+3B)x^2 + (B+3C)x + (A+C) Equating coefficients: For x^2: A+3B = 5 For x: B+3C = 2 Constant: A+C = 1 Substitute A = (4)/(5) into A+C=1: (4)/(5) + C = 1 C = 1 - (4)/(5) = (1)/(5) So, C = (1)/(5). Substitute C = (1)/(5) into B+3C=2: B + 3((1)/(5)) = 2 B + (3)/(5) = 2 B = 2 - (3)/(5) = (10)/(5) - (3)/(5) = (7)/(5) So, B = (7)/(5). Step 4: Write the final partial fraction decomposition. Substitute the values of A, B, and C back into the partial fraction form: (5x^2 + 2x + 1)/((3x+1)(1+x^2)) = (4)/(5)3x+1 + (7)/(5)x+(1)/(5)1+x^2 This can be written as: (4)/(5(3x+1)) + (7x+1)/(5(1+x^2)) That's 2 down. 3 left today — send the next one.