Chords and pieces of a circle
Mark n points on a circle and draw all the chords: how many crossings, and how many pieces does the circle fall into? A problem where the first answers 1, 2, 4, 8, 16 mislead.
How to solve it
Mark $n$ points on a circle and join every pair: the chords cut the circle into pieces. For $n = 1, 2, 3, 4, 5$ there are $1, 2, 4, 8, 16$ pieces — powers of two, it seems. But for $n = 6$ there are $31$. The right count goes through binomial coefficients: every crossing of chords is a set of four points.
Step by step
- There are as many chords as pairs of points: $\binom{n}{2}$.
- There are as many crossings inside as sets of four points: $\binom{n}{4}$. Four points on a circle give exactly one crossing, of the diagonals of their quadrilateral.
- Each chord and each crossing adds one piece: $1 + \binom{n}{2} + \binom{n}{4}$ pieces.
- The inverse problem: try $n = 1, 2, 3, \dots$ until the formula gives the number.
Common mistakes
- Continuing the pattern $1, 2, 4, 8, 16$: six points give $31$ pieces, not $32$.
- Counting crossings as pairs of chords: there are more pairs of chords than crossings, since not all chords cross inside the circle.
- Forgetting the one — the circle itself before the first chord.
- Confusing $\binom{n}{4}$ with $\frac{n^4}{4}$: $\binom{n}{4} = \frac{n(n - 1)(n - 2)(n - 3)}{24}$.