There are m, n and r points in 3 parallel (different lines). How many triangles can be formed by those points? There are m, n and r points in 3 parallel (different lines). Supposing that when taking one point from each line they're never aligned. How many triangles can be formed by those points?
 A: (for the purposes of this answer I will assume orientation is unimportant, i.e. triangle xyz is the same as triangle xzy)
Label the lines as $A$, $B$, and $C$, with points along each as $a_1,a_2,\dots,a_m$, $b_1,b_2,\dots,b_n$, and $c_1,c_2,\dots,c_r$.
According to the problem statement, for every $i,j,k$ you have $a_i,b_j,c_k$ are not colinear.
What constitutes enough information to form a triangle then?  Any three points out of the $(m+n+r)$ total points will define a unique triangle unless they are colinear.  The only way they could be colinear however is that they all lie on the same line.
So.  How many ways are there to choose 3 points (with order unimportant)?


*

*There are $m+n+r$ total points, and we want to choose 3 of them.

*$\binom{m+n+r}{3}$


How many of these groups of 3 points were in fact colinear?


*

*Either all three come from line $A$, all three come from line $B$, or all three come from line $C$

*$\binom{m}{3} + \binom{n}{3} + \binom{r}{3}$


For a total of:
$$\binom{m+n+r}{3} - \binom{m}{3}-\binom{n}{3}-\binom{r}{3}$$
