Solution: 2015 Winter Final - 4

Author: Michiel Smid

Question

What is the coefficient of $x^{4}y^{11}$ in the expansion of $(2x - 7y)^{15}$?

(a)

$- {15 \choose 11} \cdot 2^{11} \cdot 7^4$

(b)

${15 \choose 11} \cdot 2^{11} \cdot 7^4$

(c)

$- {15 \choose 11} \cdot 2^4 \cdot 7^{11}$

(d)

${15 \choose 11} \cdot 2^4 \cdot 7^{11}$

$ {(2x-7y)}^{15} $

$= \sum_{k=4}^{15} \binom{15}{k} {(2x)}^{k} {(-7y)}^{15-k} $

$ = \binom{15}{4} 2^{4} {(-7)}^{11} x^4 y^{11}$

$ = - \binom{15}{4} 2^{4} {(7)}^{11} $