What is the coefficient of $x^3 y^4$ in the expansion of $ (2x-y+5)^8$ I was thinking of doing $\binom{8}{4}$ but not sure if right. 

Notice, the binomial expansion  $$(2x-y+5)^8=^{8}C_0(2x-y)^85^0+\color{red}{^{8}C_1(2x-y)^75^1}+^{8}C_2(2x-y)^65^2+^{8}C_3(2x-y)^55^3+^{8}C_4(2x-y)^45^4+\dots +^{8}C_8(2x-y)^05^8$$
In the given term $x^3y^4$, the sum of powers $3+4=7$ hence in the above expansion there is only one term $^{8}C_1(2x-y)^75^1$ which has sum of powers $7$ hence using binomial expansion of $^{8}C_1(2x-y)^75^1$ as follows $$^{8}C_1(2x-y)^75^1=\color{blue}{5(^{8}C_1)}[^{7}C_0(2x)^7(-y)^0+^{7}C_1(2x)^6(-y)^1+^{7}C_2(2x)^5(-y)^2+^{7}C_3(2x)^4(-y)^3+\color{red}{^{7}C_4(2x)^3(-y)^4}\ldots +^{7}C_7(2x)^0(-y)^7]$$
Hence, the coefficient of $x^3y^4$ of $5(^{8}C_1)(^{7}C_4)(2x)^3(-y)^4$ is given as $$=5(^{8}C_1)(^{7}C_4)(2)^3(-1)^4$$
$$=5(8)(35)(8)(1)=\color{red}{11200}$$
