$1 \cdot 19$ For a given $n \in N$, find the number of all different natural number triples whose sum is $6 n$.



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[Solution] Let $(x, y, z)$ be a triplet of natural numbers such that $x+y+z=6n$ and $x \leqslant y \leqslant z$. When $k=1,2, \cdots, n$, all triplets $(x, y, z)$ for which $x=2k-1$ are:
$$
\begin{array}{l}
(2k-1, 2k-1, 6n-4k+2), (2k-1, 2k, 6n-4k+1), \\
\cdots, (2k-1, 3n-k, 3n-k+1),
\end{array}
$$

All triplets $(x, y, z)$ for which $x=2k$ are:
$$
\begin{array}{l}
(2k, 2k, 6n-4k), (2k, 2k+1, 6n-4k-1), \\
\cdots, (2k, 3n-k, 3n-k).
\end{array}
$$
The number of triplets in (1) and (2) is
$$
\begin{aligned}
S_{k} & =[(3n-k)-(2k-2)]+[(3n-k)-(2k-1)] \\
& =6(n-k)+3 .
\end{aligned}
$$

Summing up, the total number of all triplets is
$$
S=\sum_{k=1}^{n} S_{k}=\sum_{k=1}^{n}[6(n-k)+3]=6n^2-3n(n+1)+3n=3n^2 \text{. }
$$