Summation of Products of n Numbers taken m at a time with Repetitions/Inverse Formula/Examples/Degree 4

Example of Summation of Products of n Numbers taken m at a time with Repetitions: Inverse Formula
Let $a, b \in \Z$ be integers such that $b \ge a$.

Let $U$ be a set of $n = b - a + 1$ numbers $\set {x_a, x_{a + 1}, \ldots, x_b}$.

Let $m \in \Z_{>0}$ be a (strictly) positive integer.

Let:

That is, $h_m$ is the product of all $m$-tuples of elements of $U$ taken $m$ at a time.

For $r \in \Z_{> 0}$, let:
 * $S_r = \ds \sum_{j \mathop = a}^b {x_j}^r$

Then:
 * $S_4 = 4 h_4 - 4 h_1 h_3 - 2 {h_2}^2 + 4 {h_1}^2 h_1 - {h_1}^4$

Proof
From Summation of Products of n Numbers taken m at a time with Repetitions: Inverse Formula:
 * $S_m = \ds \sum_{k_1 \mathop + 2 k_2 \mathop + \mathop \cdots \mathop + m k_m \mathop = m} \paren {-1}^{k_1 + k_2 + \cdots + k_m - 1} \dfrac {m \paren {k_1 + k_2 + \cdots + k_m - 1}! } {k_1! \, k_2! \, \cdots k_m!} {h_1}^{k_1} {h_2}^{k_2} \cdots {h_m}^{k_m}$

where:
 * $k_1 + 2 k_2 + \cdots + m k_m = m$

We need to find all sets of $k_1, k_2, k_3, k_4 \in \Z_{\ge 0}$ such that:


 * $k_1 + 2 k_2 + 3 k_3 + 4 k_4 = 4$

Thus $\tuple {k_1, k_2, k_3, k_4}$ can be:


 * $\tuple {4, 0, 0, 0}$
 * $\tuple {2, 1, 0, 0}$
 * $\tuple {0, 2, 0, 0}$
 * $\tuple {1, 0, 1, 0}$
 * $\tuple {0, 0, 0, 1}$

Hence: