Find the coefficient of x^4 y^5 in the expansion of each of the following expres

Question

Find the coefficient of x^4 y^5 in the expansion of each of the following expressions. (a) (x + y)^9. (b) (x + y)^7. (c) (4x + 5y)^9. (d) (x - 3y)^9. Fill in the next three rows of Pascal's triangle using Pascal's identify, and verify Corollaries 1 and 2 for these rows. Verify Pascal's identify explicitly for n = 10, k = 6. Verify the identify in Example 2 explicitly for n = 10, k = 6. Expand (x + y + z)^3 by thinking combinatorially, not by multiplying it out and combining like terms. Find the coefficient of x y^3 z^2 in the expansion of (x + y + z)^6. The multinomial theorem states that the coefficient of x_1^n_1, x_2^n_2...x_ m^n_ m in the expansion of (x_1 + x_2 + ... + x_ m)^n, where n_1 + n_2 + ... + n_ m = n, is given by n!/n_1! n_2!...n_ m!. It is sometimes denoted C(n: n_1, n_2, ...n_ m). (a) Show that this specializes to the binomial theorem when m = 2. (b) Prove this result. (c) Use this result to find the coefficient of a^6 b^2 d^2 in the expansion of (2a + 3b - c - 2d)^10. Find closed forms (simple expressions involving n) for the following sums. (a) sigma_ k = 0^n C(n, k) 2^k. (b) sigma_ k = 0^n C(n, k) (-2)^k. Note that apparently every entry in the p^th row of Pascal's triangle, except the 1's, is divisible by p as long as p is a prime number. Prove that this statement is always true. Explain why the numbers C(n, 2) should be called "triangular numbers" by looking at triangular arrangements of dots. Prove that C(n, 0) + C(n + 1, 1) + C(n + 2, 2) + .. + C(2n, n) + C(2n + 1, n + 1). Consider the following situation: George and Martha invite n married couples to their house for a party. They need to choose two people from among them

Explanation / Answer

2. The co-efficient of xayb in the expansion of (px+qy)a+b is paqb(a+b)Ca or paqb(a+b)Cb

(a) (x+y)9: The co-efficient of x4y5 is 9C4

(b) (x+y)7: Since 4+5=9, the co-efficient is 0.

(c) (4x+5y)9: The co-efficient is 44*55*9C4

(d) (x-3y)9: The co-efficient is -35*9C4

6. (x+y+z)3

Combinatorially expanding,

= (3C0) x3y0z0 + (3C1) x2y1z0 + (3C1) x2y0z1 + (3C2) x1y2z0 + 2*(3C2) x1y1z1 + (3C2) x1y0z2 + (3C3) x0y3z0 + (3C2)x0y2z1 + (3C2) x0y1z2 + (3C3) x0y0z3

= x3 + 3x2y + 3x2z + 3xy2 + 6xyz + 3xz2 + y3 + 3y2z + 3yz2 + z3

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