Find the coefficient of x^2y^3 in the binomial expansion of (x-2y)^5.

Answer 1

term(r+1) = C(5,r) x^(5-r) (-2y)^r

so x^(5-r) = x^2
5-r = 2
r = 3

so the term is C(5,3) x^2 (-2y)^3
= 10 x^2 y^3 (-8)
= -80x^2y^3

or

you could just expand it
(x-2y)^5
= x^5 + 5(x^4)(-2y) + 10(x^3)(-2y)^2 + 10(x^2)(-2y)^3 + ...