49**. (Cayley's Problem). How many convex $k$-gons are there, all vertices of which coincide with the vertices of a given convex $n$-gon, and all sides are its diagonals?
49. For there to exist at least one such $k$-gon, $n$ must be no less than $2k$ (since each pair of vertices of the $k$-gon must be separated by at least one vertex of the $n$-gon).

Let the vertices of the $n$-gon be denoted by $A_{1}, A_{2}, \ldots, A_{n-1}, A_{n}$, and let us count how many $k$-gons, satisfying the conditions of the problem and having one of their vertices at $A_{n-1}$, exist. Suppose the other $k-1$ vertices of some such $k$-gon are the points $A_{i_{1}}, A_{i_{2}}, \ldots, A_{i_{k-1}}$. The numbers $i_{1}, i_{2}, \ldots, i_{k-1}$ are all between 1 and $n-3$ and must satisfy the conditions

$$
i_{2}-i_{1} \geqslant 2, i_{3}-i_{2} \geqslant 2, \ldots, i_{k-1}-i_{k-2} \geqslant 2
$$

Now consider $k-1$ numbers

$$
j_{1}=i_{1}, j_{2}=i_{2}-1, j_{3}=i_{3}-2, \ldots, j_{k-1}=i_{k-1}-(k-2)
$$

From the inequalities that the numbers $i_{1}, i_{2}, \ldots, i_{k-1}$ must satisfy, it follows that for the numbers $j_{1}, j_{2}, \ldots, j_{k-1}$, the inequalities

$1 \leqslant j_{1}<j_{2}<j_{3}<\ldots<j_{k-1} \leqslant(n-3)-(k-2)=n-k-1$.

Conversely, if $j_{1}, j_{2}, \ldots, j_{k-1}$ are arbitrary distinct numbers between 1 and $n-k-1$ and arranged in increasing order, then the numbers $i_{1}=j_{1}$, $i_{2}=j_{2}+1, \ldots, i_{k-1}=j_{k-1}+(k-2)$ will satisfy the inequalities

$$
i_{1} \geqslant 1, i_{2}-i_{1} \geqslant 2, \ldots, i_{k-1}-i_{k-2} \geqslant 2, i_{k-1} \leqslant n-3
$$

and, consequently, the $k$-gon $A_{i_{1}} A_{i_{2}} \ldots A_{i_{k-1}} A_{n-1}$ will be one of the desired $k$-gons having one of its vertices at $A_{n-1}$. Therefore, the number of the desired $k$-gons having one of their vertices at $A_{n-1}$ is equal to the number of groups of $k-1$ positive integers not exceeding $n-k-1$, i.e., it is equal to $C_{n-k-1}^{k-1}$.

Now it is easy to calculate the total number of $k$-gons satisfying the conditions of our problem. Since the number of $n$-gons having one of their vertices at a specific vertex of the $n$-gon is the same for all vertices, multiplying $C_{n-k-1}^{k-1}$ by $n$ (i.e., summing the number of $k$-gons having one of their vertices at a specific vertex of the $n$-gon over all $n$ vertices), we will count each of the $k$-gons satisfying the conditions of the problem exactly $k$ times (a $k$-gon has $k$ vertices). Therefore, the number of different $k$-gons is

$$
\frac{n}{k} C_{n-k-1}^{k-1}=\frac{n(n-k-1)!}{k!(n-2k)!}
$$