Solution: 2015 Winter Final - 4

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

$ = \sum_{k=0}^{15} \binom{15}{k} {(2x)}^{n-k} {(-7y)}^{k} $

We only consider $k=11$, as it results in $y^{11}$.

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

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

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

Thus, the coefficient of $ x^{4}y^{11} $ in the expansion of $ {(2x-7y)}^{15} $ is $ - \binom{15}{11} \cdot {2}^{4} \cdot {7}^{11} $