In space, there are 10 different points, none of which are in the same plane. Among them, 5 are on one line, another 3 are on a line parallel to this one, and the remaining 2 points define a line parallel to the previous ones. How many (real) triangles do these points define?
I. solution. Let's denote the three parallel lines containing 5, 3, and 2 points respectively as $e$, $f$, and $g$. Since no three vertices of a real triangle lie on the same line, at most two of the vertices of the sought triangles can be points on the same one of our lines.

Thus, we can categorize the sought triangles into 4 groups:

I. All three vertices come from different lines. The selection of these can start in 5 ways by choosing a point from the line $e$, each of these can be continued in 3 ways by choosing a point from the line $f$, and the resulting $5 \cdot 3$ point pairs can be completed in $5 \cdot 3 \cdot 2=30$ ways by considering the possibilities of the points on $g$.

II. Two vertices are from $e$ - that is, one side of the triangle - and the third is from $f$ or $g$. In this case, the first point from $e$ can be chosen in 5 ways, and the second from the remaining 4 points in 4 ways, so the $5 \cdot 4=20$ selections, however, represent the same pair of points for the side of the triangle, as we considered $AB$ and $BA$ as well. The side from $e$ can thus be chosen in $20: 2=10$ ways, and in each case, the opposite vertex can be chosen from the $3+2=5$ points of $f$ and $g$ together, resulting in 5 triangles, a total of $10(3+2)=50$ triangles.

III. Two vertices are from $f$. These can be combined in $(3 \cdot 2): 2=3$ ways (alternatively: from the 3 points, we either do not choose the first, or the second, or the third as a vertex), and the number of different triangles is $3(5+2)=21$.

IV. Two vertices are from $g$, that is, both $g$-line points are chosen, and the third from the remaining $5+3=8$. Thus, there are 8 different such triangles.

No triangle belongs to two groups, so in total, there are $30+50+21+8=109$ possible triangles.

II. solution. First, we think of all the triplets of points that can be formed from the points, then we remove those that do not form a real triangle because their three points lie on the same line.

The first point of the triplet can be chosen in 10 ways, each of these starting points can be continued in 9 ways by choosing from the remaining 9 points, and finally, the remaining points offer 8 possibilities for completion. Thus, we have thought of $10 \cdot 9 \cdot 8=720$ triplets of points.

However, not all of these thought triplets represent different triangles, since the vertices of any $ABC$ triangle can be listed in 6 ways:

$$
ABC, \quad ACB, \quad BAC, \quad BCA, \quad CAB, \quad CBA
$$

Accordingly, the number of different triplets is $720: 6=120$.

With the same reasoning, we get the number of triplets whose all three elements lie on the line $e$: $(5 \cdot 4 \cdot 3): 6=10$, similarly, on the line $f$ there is $(3 \cdot 2 \cdot 1): 6=1$ triplet to be excluded, and on $g$ there is none.

The remaining - and now suitable - triplets, that is, the number of real triangles is $120-(10+1)=109$.