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## Here's the Solution to this Question

a)

$(x+y)^6=x^6+6x^5y+15x^4y^2+20x^3y^3+15x^2y^4$

$+6xy^5+y^6$

b)

$(x+y)^4=x^4+4x^3y+6x^2y^2+4xy^3+y^4$

31)

$\dbinom{25}{12}=\dfrac{25!}{12!(25-12)!}=5200300$

32)

$(-1)^9\dbinom{19}{10}(2)^{10}=-\dfrac{19!}{10!(19-10)!}(1024)$

$=-92378(1024)=-94595072$

33)

$(-1)^{99}\dbinom{200}{101}(2)^{101}(3)^{99}$

$=-\dfrac{200!}{101!(200-101)!}(2)^{101}(3)^{99}$

a)

$(x+y)^6=x^6+6x^5y+15x^4y^2+20x^3y^3+15x^2y^4$

$+6xy^5+y^6$

b)

$(x+y)^4=x^4+4x^3y+6x^2y^2+4xy^3+y^4$

31)

$\dbinom{25}{12}=\dfrac{25!}{12!(25-12)!}=5200300$

32)

$(-1)^9\dbinom{19}{10}(2)^{10}=-\dfrac{19!}{10!(19-10)!}(1024)$

$=-92378(1024)=-94595072$

33)

$(-1)^{99}\dbinom{200}{101}(2)^{101}(3)^{99}$

$=-\dfrac{200!}{101!(200-101)!}(2)^{101}(3)^{99}$

a)

$(x+y)^6=x^6+6x^5y+15x^4y^2+20x^3y^3+15x^2y^4$

$+6xy^5+y^6$

b)

$(x+y)^4=x^4+4x^3y+6x^2y^2+4xy^3+y^4$

31)

$\dbinom{25}{12}=\dfrac{25!}{12!(25-12)!}=5200300$

32)

$(-1)^9\dbinom{19}{10}(2)^{10}=-\dfrac{19!}{10!(19-10)!}(1024)$

$=-92378(1024)=-94595072$

33)

$(-1)^{99}\dbinom{200}{101}(2)^{101}(3)^{99}$

$=-\dfrac{200!}{101!(200-101)!}(2)^{101}(3)^{99}$

a)

$(x+y)^6=x^6+6x^5y+15x^4y^2+20x^3y^3+15x^2y^4$

$+6xy^5+y^6$

b)

$(x+y)^4=x^4+4x^3y+6x^2y^2+4xy^3+y^4$

31)

$\dbinom{25}{12}=\dfrac{25!}{12!(25-12)!}=5200300$

32)

$(-1)^9\dbinom{19}{10}(2)^{10}=-\dfrac{19!}{10!(19-10)!}(1024)$

$=-92378(1024)=-94595072$

33)

$(-1)^{99}\dbinom{200}{101}(2)^{101}(3)^{99}$

$=-\dfrac{200!}{101!(200-101)!}(2)^{101}(3)^{99}$