Three neighbours in a row of Pascal's triangle, ⁿC_(r - 1), ⁿCr and ⁿC_(r + 1), are in the ratio 1…

$^{n}C_{r} = {}^{n}C_{n - r}$; if $^{n}C_{a} = {}^{n}C_{b}$ then $a = b$ or $a + b = n$; Pascal's rule $^{n}C_{r} + {}^{n}C_{r - 1} = {}^{n + 1}C_{r}$; and $\dfrac{^{n}C_{r}}{^{n}C_{r - 1}} = \dfrac{n - r + 1}{r}$. With $n$ points, no three in a line, there are $^{n}C_{2}$ lines and $^{n}C_{3}$ triangles. A polygon of $n$ sides has $^{n}C_{2} - n$ diagonals.