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Answer 1 · 2018-01-26T10:32:46

#-101376#

Looking for the coefficient of #a^5b^7# in the expansion of #(a-2b)^12#.

The binomial theorem says #(x+y)^n=sum_(k=0)^n((n),(k))x^(n-k)*y^k#

So #(a-2b)^12 = sum_(k=0)^12((12),(k))a^(12-k)*(-2b)^k#.

For the term we're seeking, we need the term when #k=7#:

#((12),(7))a^(12-7)*(-2b)^7#

#=792a^5(-2)^7b^7#

#=-792*128a^5b^7#

#=-101376a^5b^7#

so the coefficient is #-101376#.

Note:
#((12),(7))=(12!)/(7!(12-7)!) #
#= ((12)(11)(10)(9)(8))/((5)(4)(3)(2)(1))#
# = (11)(9)(8) =792#