How to determine the coefficient of binomial Suppose I have $\left(x-2y+3z^{-1}\right)^4$
How to determine coefficient binom of $xyz^{-2}$?
I've tried using trinominal expansion like this:
$\displaystyle \frac{4!}{1!1!1!} (1)^1 (-2)^1 (3)^2$
 A: $$\begin{align*}
\left(x-2y+3z^{-1}\right)^4 =& \left[(x-2y)+3z^{-1}\right]^4\\
=& \sum_{i=0}^4 \binom4i (x-2y)^{4-i}\left(3z^{-1}\right)^i\\
=& \binom42 (x-2y)^2\left(3z^{-1}\right)^2 + \cdots\\
=& \binom42 \left(x^2-4xy+4y^2\right)\left(3z^{-1}\right)^2 + \cdots\\
=& \binom42 \left(-4xy\right)\left(3z^{-1}\right)^2 + \cdots\\
\end{align*}$$
Other non-$xyz^{-2}$ terms are omitted.
A: Note that
$$(a+b+c)^n = \sum_{k=0}^n {n\choose k}(a+b)^{n-k}c^k = \sum_{k=0}^n\sum_{j=0}^{n-k}{n\choose k}{{n-k}\choose j}a^{n-k-j}b^jc^k$$
So in your problem, we have that
$$(x-2y+3z^{-1})^4 = \sum_{k=0}^4\sum_{j=0}^{4-k}{4\choose k}{4-k\choose j}x^{4-k-j}(-2y)^j(3z^{-1})^k$$
Thus, we get the $xyz^{-2}$ term when $k=2$ and $j=1$.  Computing the coefficient from here should be straight forward.
A: You can do this by elementary combinatorics, think about how products of this kind are expanded. You need to find all the possible (ordered) products of $x$, $y$ and $z^{-1}$ with 4 factors that give $xyz^{-2}$. In order to obtain $xyz^{-2}$ one of the factors needs to $x$, one needs to be $y$ and two need to be $z^{-1}$, so we are looking for all the permutations of $(x,y,z^{-1},z^{-1})$. In general, there are $4!=24$ permutations of a $4$-tuple, but since two of the entries are equal, we are left with $4!/2=12$ different permutations. Now you just multiply the coefficients of these monomials with how often the desired combination appears in the expansion and obtain
$$12\cdot x\cdot(-2)y\cdot3z^{-1}\cdot3z^{-1} = -216xyz^{-2}.$$
This is the reasoning that leads to multinomial expansion in the end, but I find it a lot easier to remember how this works, instead of shooting up the multinomial formula if I only need a single coefficient.
A: The Multinomial Theorem: Let $n$ be a positive integer. For all $x_1,x_2,...,x_t$, $$(x_1+x_2+\cdots +x_t)^n=\sum {n\choose n_{1}n_{2}\cdots{n_t}}x_1^{n_1}x_2^{n_2}\cdots x_t^{n_t},$$ where the summation extends over all non-negative integral solutions $n_1,n_2,\cdots,n_t$ of $n_1+n_2+\cdots +n_t=n$. Since we are concered with the coefficient of $xyz^{-2}$ in $(x-2y+3z^{-1})^4$ we can compute this directly as ${4\choose {1}{1}{2}}(1)^1(-2)^1(3)^2$ where $n=4$, $n_1=1$, $n_2=1$, $n_3=2$, $x_1=1$, $x_2=-2$, and $x_3=3$.
