5】 Given a set of points on a plane $P=\left\{p_{1}, p_{2}, \cdots, p_{194}\right\},$ where no three points in $P$ are collinear. Divide the points in $P$ into 83 groups, with each group containing at least 3 points. Connect every pair of points within the same group, but do not connect points from different groups, to form a graph $G$. The number of triangles in $G$ is denoted as $m(G)$.
(1) Find the minimum value of $m(G)$.
(2) Let the graph that achieves the minimum $m(G)$ be $G^{\prime}$. Prove that the points in $G^{\prime}$ can be 4-colored such that $G^{\prime}$ contains no monochromatic triangles. (1994 National High School Mathematics Competition Problem)
5. (1) Since the number of grouping methods is finite, there must exist a grouping method that results in the minimum number of triangles. Note that $1994=83 \times 24+2=81 \times 24+2 \times 25$, thus, we can divide 1994 points into 83 groups, where 81 groups each have 24 points, and 2 groups each have 25 points. We prove that such a grouping results in the minimum number of triangles, i.e., $(m(G))_{\min }=81 \mathrm{C}_{24}^{3}+2 \mathrm{C}_{25}^{3}=168544$. Otherwise, for $(i, j)(i \geqslant j+2), S=\mathrm{C}_{i}^{3}+\mathrm{C}_{j}^{3}$, if we adjust to $(i-1, j+1), S^{\prime}=\mathrm{C}_{i-1}^{3}+\mathrm{C}_{j+1}^{3}$, then $S-S^{\prime}=\mathrm{C}_{i}^{3}+\mathrm{C}_{j}^{3}-\left(\mathrm{C}_{i-1}^{3}+\mathrm{C}_{j+1}^{3}\right)=\mathrm{C}_{i-1}^{2}-\mathrm{C}_{j}^{2}>0$. By making such an adjustment once, the number of triangles decreases. (2) $G^{\prime}$ consists of several independent connected graphs, so we only need to consider the coloring of two graphs $G_{1}$ and $G_{2}$ with $\left|G_{1}\right|=25$ and $\left|G_{2}\right|=24$. Furthermore, we only need to consider the coloring of graph $G_{1}$. In fact, for $\left|G_{2}\right|=24$, we can remove one point and its associated edges from the coloring of $G_{1}$. Divide 25 points into 5 groups, each with 5 points. In each group of 5 points, the edges 12, 23, 34, 45, 51 are colored with the first color; the edges 13, 35, 52, 24, 41 are colored with the second color. Then, treat the colored group of five points as a "large point", resulting in 5 "large points". For these 5 large points, color the edges between them using the above method with two additional colors, thus completing the coloring with 4 colors.