I wanted to find the number of diagonals to a dodecagon. I know the standard way is to use the formula $\frac{n(n-3)}{2}$, but I always forget the formula, so I used combinatorics. I tackled the problem from the perspective of drawing lines starting from a point. Since two adjacent points cannot create a diagonal, you have nine other dots to select from for where to draw the diagonal. Because the starting dot does not matter, this leaves us with $\binom{11}{9}$ possible drawings. Because it is possible to choose to draw to the same dot, we must exclude that possibility. Thus, there are $\binom{11}{9}-1$ diagonals to the dodecagon.
If I were to generalize this to some regular $n$-gon, then there are $\binom{n-1}{n-3}-1$ diagonals. Can this reasoning apply to any regular $n$-gon? I'd rather not memorize that formula, and this method seems to work (I tried it with $n = 6, 7, 12$).
Best Answer
Yes, both formulae are correct.
$$ \begin{align*} {{n-1}\choose{n-3}} - 1 &= \frac{(n-1)!}{(n-3)!\cdot 2!} - 1 \\ &= \frac{(n-1)(n-2)(n-3)\cdots 2 \cdot 1}{2(n-3)(n-4)\cdots 2 \cdot 1} - 1 \\ &= \frac{(n-1)(n-2)}{2} - 1 \\ &= \frac{n^2-3n+2}{2} - \frac{2}{2} \\ &= \frac{n(n-3)}{2} \end{align*} $$