What is the coefficient of $x^3 y^4$ in the expansion of $ (2x-y+5)^8$

Question

I was thinking of doing $\binom{8}{4}$ but not sure if right.

Answer 1

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 $$=5(^{8}C_1)(^{7}C_4)(2)^3(-1)^4$$ $$=5(8)(35)(8)(1)=\color{red}{11200}$$

Answer 2

HINT:

The $r,0\le r\le8$th term of $$(2x-y+5)^8=\{(2x+5)-y\}^8$$ is $$\binom8ry^{8-r}(2x+5)^{8-r}(-y)^r$$

So, we need $r=4$

So, we focus on $$\binom84y^{8-4}(2x+5)^{8-4}(-y)^4$$

Now the $n,0\le n\le4$th term of $$(2x+5)^4$$ is $$\binom4n(2x)^{4-n}5^n$$

We need $4-n=3$

Answer 3

One has $$(2x-y+5)^8 = \sum_{i=0}^8 {8 \choose i} (2x-y)^i 5^{8-i} = \sum_{i=0}^8 {8 \choose i} 5^{8-i}\sum_{j=0}^i {i\choose j} (2x)^jy^{i-j}(-1)^{i-j}.$$

Then, one gets the coefficient of $x^3y^4$ is $${8 \choose 7}5^{8-7}{7\choose 3}2^3.$$

Answer 4

Pre-expansion, there are $8$ factors of $2x - y + 5$.

From those $8$ factors, choose the $3$ that contribute to the $x^3$, from the remaining $5$ factors, choose the $4$ that will contribute to $y^4$. There is only 1 factor left so it chooses itself.

Post-expansion, that's ${8 \choose 3}{5 \choose 4}{1 \choose 1}$ terms with $x^3y^4$, and taking into account the initial coefficients of $\{2, -1, 5\}$, the final coefficient is:

$${8 \choose 3}{5 \choose 4}2^3(-1)^45^1 = 11200$$

Answer 5

You can also solve it using the multinomial theorem.