How do you find the coefficient of #x^5# in the expansion of #(x-3)^7#?

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Answer 1 · 2016-12-10T14:56:57

189

for integers the binomial expansion is

#(x+y)^n=x^n+^nC_1x^(n-1)y+^nC_2n^(n-2)y^2+....^nC_kx^(n-k)y^k+.....y^n#

where#" "^nC_k=(n!)/(k!(n-k)!)#

for #" "(x-3)^7#

the term including #" "x^5#

#" "^7C_2x^5(-3)^2#

#=(7!)/(2!(5!))xx9#

#=((7xx6xxcancel(5!))/(2xxcancel(5!)))xx9#

#=21xx9=189#