How many triangles are determined by $n$ points, if (1) no 3 points lie on the same line and if (2) $p_{1}$ points lie on one line, and $p_{2}$ points lie on another line?
$$
1^{\circ} \cdot\binom{n}{3} . \quad 2^{\circ} \cdot\binom{n}{3}-\binom{p_{1}}{3}-\binom{p_{2}}{3} .
$$

Number of solutions: 24.