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 $-13824\cdot1680=-23~224~320$.
Binomial theorem states that the $k^{th}$ term (if we start counting from $0$) of the expansion is $$\binom{10}{k}\cdot 4^k \cdot x^{10-k}$$
The coefficient of any term is therefore going to be $\binom{10}{k}\cdot 4^k$. We will try to maximize this value. I will do this by calculator:
$k=10, 4^{10}= 1048576, \binom{10}{10}=1 \to 1048576$
$k=9, 4^{9}= 262144, \binom{10}{9}=10 \to 2621440$
$k=8, 4^{8}= 65536, \binom{10}{8}=45 \to 2949120$
$k=7, 4^{7}= 16384, \binom{10}{7}=120 \to 1966080$
$k=6, 4^{6}= 4096, \binom{10}{6}=210 \to 860160$
$k=5, 4^{5}= 1024, \binom{10}{5}=252 \to 258048$
After this, both the binomial coefficients and the power of $4$ will only decrease. Therefore, the largest coefficient is at $k=8$, and is 2949120.
Best Answer
Hint. Note that $\sum_{k=1}^{20} k=210$ and $$(x-1)(x^2 - 2)(x^3-3)\cdots(x^{20} - 20)=x^{210} \left(1-\frac{1}{x}\right) \left(1-\frac{2}{x^2}\right) \left(1-\frac{3}{x^3}\right)\cdots\left(1-\frac{20}{x^{20}}\right).$$ Now consider the integer partitions with distinct parts of $210-203=7$: $$7,\quad 6+1,\quad 5+2,\quad 4+3,\quad 4+2+1.$$