The exercise says:
In the expansion $(2x+3y-4z+w)^9$, find the coefficient of
$x^3y^2z^3$.
The formula to find the coefficient of $x_1^{r_1}x_2^{r^2}\dots x_k^{r_k}$ in $(x_1+x_2+\dots+x_k)^n$ is:
$$\frac{n!}{r_1!r_2!\dots r_k!}$$
Am I supposed to multiply the result determined by the above formula by $2$, $3$ and $4$? What to do when $x$ and $y$ has coefficients?
Best Answer
Every term in $(2x+3y-4z+w)^9$ is a product of nine factors, each of which is one of the four terms in parentheses. Thus, before you collect like terms each factor will have the form $(2x)^i(3y)^j(-4z)^kw^\ell$, where $i+j+k+\ell=9$. Since the exponents in $x^3y^2z^3$ add up to only $8$, not $9$, there is no such term in the product, and its coefficient is $0$.
If you actually meant the coefficient of $x^3y^3z^3$, each such term must arise as the product of three factors of $2x$, three of $3y$, and three of $-4z$, so it must be $(2x)^3(3y)^3(-4z)^3=2^33^3(-4)^3x^3y^3z^3$, with a coefficient of $2^33^3(-4)^3=-13824$. Your formulat tells you that there are
$$\frac{9!}{3!3!3!}=1680$$
such terms, so the total coefficient of $x^3y^3z^3$ is $-13824\cdot1680=-23~224~320$.