11.6. On a circle, 20 points are marked. How many triples of chords with endpoints at these points exist such that each chord intersects each other (possibly at the endpoints)?
Answer: 156180.

Solution. The ends of the sought chords can be 3, 4, 5, or 6 points. Let's consider these cases.

1) The ends of the chords are 3 points (see Fig. 11.6a). They can be chosen in $C_{20}^{3}$ ways. Each triplet of points can be connected by chords in a unique way.

![](https://cdn.mathpix.com/cropped/2024_05_06_c9ae9b97d8e763e17998g-3.jpg?height=309&width=390&top_left_y=436&top_left_x=114)

Fig. 11.6a

![](https://cdn.mathpix.com/cropped/2024_05_06_c9ae9b97d8e763e17998g-3.jpg?height=360&width=355&top_left_y=383&top_left_x=531)

Fig. 11.6b

![](https://cdn.mathpix.com/cropped/2024_05_06_c9ae9b97d8e763e17998g-3.jpg?height=360&width=351&top_left_y=383&top_left_x=912)

Fig. 11.6c

![](https://cdn.mathpix.com/cropped/2024_05_06_c9ae9b97d8e763e17998g-3.jpg?height=363&width=330&top_left_y=382&top_left_x=1295)

Fig. 11.6d

![](https://cdn.mathpix.com/cropped/2024_05_06_c9ae9b97d8e763e17998g-3.jpg?height=342&width=381&top_left_y=401&top_left_x=1660)

Fig. 11.6e

2) The ends of the chords are 4 points. There are two possible configurations of the chords (see Fig. 11.6b, c). Four points can be chosen in $C_{20}^{4}$ ways. For each set of four points, there are 8 ways to connect them with chords.
3) The ends of the chords are 5 points. In this case, exactly two chords share a common vertex, and the third chord connects the remaining two points (see Fig. 11.6d). Five points can be chosen in $C_{20}^{5}$ ways. For each set of five points, there are five ways to draw the chords (one for each point where two chords meet).
4) The ends of the chords are 6 points (see Fig. 11.6e). Six points can be chosen in $C_{20}^{6}$ ways. For each set of six points, there is a unique way to draw the chords, as the chords must intersect pairwise at internal points.

Thus, the total number of ways to draw the chords is

$$
\begin{aligned}
C_{20}^{3}+C_{20}^{4} \cdot 8 & +C_{20}^{5} \cdot 5+C_{20}^{6}=\frac{20 \cdot 19 \cdot 18}{2 \cdot 3}+\frac{20 \cdot 19 \cdot 18 \cdot 17 \cdot 8}{2 \cdot 3 \cdot 4}+\frac{20 \cdot 19 \cdot 18 \cdot 17 \cdot 16 \cdot 5}{2 \cdot 3 \cdot 4 \cdot 5}+\frac{20 \cdot 19 \cdot 18 \cdot 17 \cdot 16 \cdot 15}{2 \cdot 3 \cdot 4 \cdot 5 \cdot 6}= \\
& =\frac{20 \cdot 19 \cdot 18}{2 \cdot 3}\left(1+34+\frac{17 \cdot 16}{4}+\frac{17 \cdot 16 \cdot 15}{4 \cdot 5 \cdot 6}\right)=10 \cdot 19 \cdot 6 \cdot(35+68+34)=10 \cdot 19 \cdot 6 \cdot 137=156180
\end{aligned}
$$

The answer can be left in the form $C_{20}^{3}+C_{20}^{4} \cdot 8+C_{20}^{5} \cdot 5+C_{20}^{6}$.

## Grading Criteria:

+ Correct answer and a fully justified solution

$\pm$ All cases are correctly analyzed, but there is an arithmetic error in the final calculation

干 Some individual cases are correctly analyzed

- Only the answer is provided