Geometric Counting

From $n$ points, no three collinear: lines $^nC_2$, triangles $^nC_3$, quadrilaterals $^nC_4$. If $k$ points are collinear: subtract their degenerate selections — lines $^nC_2-^kC_2+1$; triangles $^nC_3-^kC_3$; quadrilaterals $^nC_4-^kC_4-^kC_3\cdot{}^{n-k}C_1$ (a second degenerate class: exactly THREE of the four vertices collinear). Diagonals of an $n$-gon: $^nC_2-n=\dfrac{n(n-3)}{2}$. Parallelograms from $m$ and $n$ parallel lines: $^mC_2\cdot{}^nC_2$. Max intersection points: $n$ lines: $^nC_2$ (each pair meets once). $n$ circles: $2\cdot{}^nC_2=n(n-1)$ — each pair of circles meets TWICE, so 5 circles give 20 and 10 circles give 90; forgetting the factor 2 is the recorded common error. $n$ lines and $m$ circles: $^nC_2+2\cdot{}^mC_2+2nm$. Triangles from the vertices of a regular $n$-gon by shared sides: exactly two sides in common (three consecutive vertices) — $n$; exactly one side — $n(n-4)$; no side in common — $^nC_3-n-n(n-4)=\dfrac{n(n-4)(n-5)}{6}$ (for an octagon: 16).