5(a) Express \( \frac{x^{2}-11}{(x+2)^{2}(3 x-1)} \) in partial fractions (b) Calculate the coefficient of \( x^{3} y^{4} \) in \( (2 x-3 y)^{7} \) 6(a) How many 2-digit numbers can be formed from the three numerals \( 2,5,6 \) ? (b) In how many ways can the President form a cabinet of 6 men and 3 women from (c) Show that \( r!^{n} C_{r}={ }^{n} P_{r} \)

Let's solve each part of the problem step by step. ### 5(a) Partial Fraction Decomposition We want to express the function \( \frac{x^{2}-11}{(x+2)^{2}(3x-1)} \) in partial fractions. The denominator can be factored as \( (x+2)^{2}(3x-1) \). The form of the partial fraction decomposition will be: \[ \frac{x^{2}-11}{(x+2)^{2}(3x-1)} = \frac{A}{x+2} + \frac{B}{(x+2)^{2}} + \frac{C}{3x-1} \] Multiplying through by the denominator \( (x+2)^{2}(3x-1) \) gives: \[ x^{2} - 11 = A(3x-1)(x+2)^{2} + B(3x-1) + C(x+2)^{2} \] Now we need to expand the right-hand side and collect like terms. 1. **Expand \( A(3x-1)(x+2)^{2} \)**: \[ (x+2)^{2} = x^{2} + 4x + 4 \] \[ A(3x-1)(x^{2} + 4x + 4) = A(3x^{3} + 12x^{2} + 12x - x^{2} - 4x - 4) = A(3x^{3} + 11x^{2} + 8x - 4) \] 2. **Expand \( C(x+2)^{2} \)**: \[ C(x^{2} + 4x + 4) = Cx^{2} + 4Cx + 4C \] 3. **Combine all terms**: \[ x^{2} - 11 = (3A)x^{3} + (11A + C)x^{2} + (8A + 4C + B)x + (-4A + 4C) \] Now, we equate coefficients from both sides: - For \( x^{3} \): \( 3A = 0 \) → \( A = 0 \) - For \( x^{2} \): \( 11A + C = 1 \) → \( C = 1 \) (since \( A = 0 \)) - For \( x^{1} \): \( 8A + 4C + B = 0 \) → \( 4 + B = 0 \) → \( B = -4 \) - For the constant term: \( -4A + 4C = -11 \) → \( 4 = -11 \) (which is consistent) Thus, we have: \[ A = 0, \quad B = -4, \quad C = 1 \] The partial fraction decomposition is: \[ \frac{x^{2}-11}{(x+2)^{2}(3x-1)} = \frac{-4}{(x+2)^{2}} + \frac{1}{3x-1} \] ### 5(b) Coefficient of \( x^{3} y^{4} \) in \( (2x - 3y)^{7} \) To find the coefficient of \( x^{3} y^{4} \) in \( (2x - 3y)^{7} \), we can use the binomial theorem: \[ (a + b)^{n} = \sum_{k=0}^{n} \binom{n}{k} a^{n-k} b^{k} \] Here, \( a = 2x \), \( b = -3y \), and \( n = 7 \). We want the term where \( (2x)^{3} \) and \( (-3y)^{4} \) appear. The term corresponding to \( k = 4 \) (since \( y \) has exponent 4) is: \[ \binom{7}{4} (2x)^{3} (-3y)^{4} \] Calculating this: 1. **Binomial coefficient**: \[ \binom{7}{4} = \frac{7!}{4!(7-4)!} = \frac{7 \times 6 \times 5}{3 \times 2 \times 1} = 35 \] 2. **Calculate the powers