All-Soviet-Union Mathematical Olympiad

Solution:
There are $30 \times 29 \times 28 / 6 = 4060$ triples in all. Let $m$ be the number of triples with $0$ or $3$ edges, and let $n$ be the number of triples with $1$ or $2$ edges. So $m + n = 4060$.

Each point is joined to $6$ others, so it is in $6 \times 5 / 2 = 15$ triples where it is joined to both the other points, and it is in $23 \times 22 / 2 = 253$ triples where it is not joined to either of the other points. So the total number of triples $(a, b, c)$, where $a$ is joined to $b$ and $c$ or not joined to $b$ or $c$ is $30(15 + 253) = 8040$.

This counts the $m$ triples $3$ times each, and the $n$ triples once each, so $3m + n = 8040$. Hence $m = 1990$.