Find the coefficient of $x^3y^2z^3$ in the expansion $(2x+3y-4z+w)^9$

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 $-13824\cdot1680=-23~224~320$.

Surely the question (although perhaps badly worded) is asking for the term in the expansion which contains $x^3y^2z^3$ as written? This term will be of the form $nx^3y^2z^3w$ for some integer $n$, so its "coefficient" is $nw$. We can expand

$(2x+3y-4z+w)^9 = ((2x+3y-4z)+w)^9$

as

$(2x+3y-4z)^9 + 9w(2x+3y-4z)^8 + \ldots$,

so $n$ is equal to 9 times the coefficient of $x^3y^2z^3$ in the expansion of $(2x+3y-4z)^8$. From Brian M. Scott's answer, we see that $n$ is equal to $$9 \times \frac{8!}{3!2!3!}$$ which I will let you calculate yourself. Also note that this is equal to $$\frac{9!}{3!2!3!1!}$$ which is the coefficient of $x^3y^2z^3w$, as expected.