44*. In how many points do the diagonals of a convex n-gon intersect if no three of them intersect at the same point?
44. First solution. Consider the diagonal $A_{1} A_{k}$ of the $n$-gon $A_{1} A_{2} A_{3} \ldots A_{n}$. On one side and the other of this diagonal, there are respectively $k-2$ vertices of the $n$-gon (vertices $A_{2}, A_{3}, A_{4}, \ldots, A_{k-1}$) and $n-k$ vertices (vertices $A_{k+1}, A_{k+2}, \ldots, A_{n}$). The diagonal $A_{1} A_{k}$ will intersect with all diagonals connecting any of the $k-2$ vertices of the first group with any of the $n-k$ vertices of the second group, i.e., with $(k-2)(n-k)$ diagonals. Therefore, all diagonals emanating from vertex $A_{1}$ (diagonals $A_{1} A_{3}, A_{1} A_{4}, \ldots, A_{1} A_{n-1}$) will intersect with other diagonals at

$$
1 \cdot(n-3)+2(n-4)+3(n-5)+\ldots+(n-3) \cdot 1
$$

points. But

$$
\begin{aligned}
& 1(n-3)+2(n-4)+3(n-5)+\ldots+(n-3) \cdot 1= \\
& =1[(n-1)-2]+2[(n-1)-3]+ \\
& \quad+3[(n-1)-4]+\ldots+(n-3)[(n-1)-(n-2)]= \\
& =(n-1)[1+2+3+\ldots+(n-3)]- \\
& \quad-[1 \cdot 2+2 \cdot 3+3 \cdot 4+\ldots+(n-3)(n-2)]= \\
& =(n-1) \frac{(n-2)(n-3)}{2}-[1 \cdot 2+2 \cdot 3+3 \cdot 4+\ldots+(n-3)(n-2)] .
\end{aligned}
$$

And since ${ }^{1}$ )

$$
\begin{aligned}
& 1 \cdot 2+2 \cdot 3+3 \cdot 4+\ldots+(n-3)(n-2)= \\
&=\frac{(n-3)(n-2)(n-1)}{3}
\end{aligned}
$$

then

$$
\begin{aligned}
& 1 \cdot(n-3)+2(n-4)+3(n-5)+\ldots+(n-3) \cdot 1= \\
& =\frac{(n-1)(n-2)(n-3)}{2}-\frac{(n-1)(n-2)(n-3)}{3}= \\
& \quad=\frac{(n-1)(n-2)(n-3)}{6} .
\end{aligned}
$$

Thus, the diagonals emanating from vertex $A_{1}$ will intersect with other diagonals at $\frac{(n-1)(n-2)(n-3)}{6}$ points. The diagonals emanating from any other vertex of the $n$-gon will intersect with other diagonals at the same number of points. However, multiplying $\frac{(n-1)(n-2)(n-3)}{6}$ by the number of vertices of the $n$-gon (which is $n$), we will count each intersection point exactly four times (in each of these points, exactly two diagonals intersect, each of which has two[^13]

see the footnote on page 179.

Similarly

$1 \cdot 2 \cdot 3+2 \cdot 3 \cdot 4+3 \cdot 4 \cdot 5+\ldots+n(n+1)(n+2)=$

$=1(1+1)(2+2)+2(2+1)(2+2)+\ldots+n(n+1)(n+2)=$ $=\left(1^{3}+2^{3}+\ldots+n^{3}\right)+3\left(1^{2}+2^{2}+\ldots+n^{2}\right)+2(1+2+\ldots+n) ;$ see the footnote on page 179 and 188.
ends in two vertices). The final number of intersection points is

$$
\frac{(n-1)(n-2)(n-3)}{6} \cdot \frac{n}{4}=\frac{n(n-1)(n-2)(n-3)}{24}
$$

Second solution. This problem can be solved much more simply if we know the formula for combinations. Indeed, consider any intersection point of diagonals. Each intersection point belongs to two diagonals. The point $B$ of intersection of diagonals $A_{1} A_{3}$ and $A_{2} A_{1}$ corresponds to a quadruple of vertices $A_{1}, A_{2}, A_{3}$, and $A_{4}$ of our $n$-gon (point $B$ is the intersection point of the diagonals of the quadrilateral $A_{1} A_{2} A_{3} A_{4}$, see Fig. 58). Conversely, each quadruple of vertices $A_{1}, A_{2}, A_{3}$, and $A_{4}$ of the $n$-gon corresponds to a unique intersection point of diagonals whose endpoints are these four vertices, namely, the intersection point of the diagonals of the quadrilateral $A_{1} A_{2} A_{3} A_{4}$. Therefore, the number of intersection points of diagonals of a convex $n$-gon is equal to the number of combinations of $n$ elements taken 4 at a time, i.e., it is equal to

$$
C_{n}^{4}=\frac{n(n-1)(n-2)(n-3)}{24}
$$