<s>[INST] How many vertices so that the number of pentagons equals the number of triangles? Question: A set of points is chosen on the circumference of a circle so that the
number of different triangles with all three vertices among the points is equal to the
number of pentagons with all five vertices in the set. How many points are there?
Solution
$$\frac{x!}{3!(x-3)!}=\frac{x!}{5!(x-5)!}\\
\frac{x(x-1)(x-2)(x-3)!}{3!(x-3)!}=\frac{x(x-1)(x-2)(x-3)(x-4)(x-5)!}{5!(x-5)!}\\
\frac{x(x-1)(x-2)}{6}=\frac{x(x-1)(x-2)(x-3)(x-4)}{120}\\
6x(x-1)(x-2)(x-3)(x-4)=120x(x-1)(x-2)$$
Hence, $x=1, x=1, x=2, x=-1, x=8$. Since a pentagon has $5$ vertices, the only plausible solution is 8. I couldn't find the solution online, just making sure.
 [/INST] Yes, this reasoning is correct. (Of course, $(n - 5)!$ is not defined for integers $\leq 4$, so we can discard those solutions outright for that reason, too.)
Alternatively, if there are $3 + 5 = 8$ vertices, then we can define a natural bijection $$\{\textrm{triangles}\} \leftrightarrow \{\textrm{pentagons}\}$$ taking a triangle, throwing out its vertices, and forming a pentagon from the remaining five, so, e.g., mapping the blue triangle to the red pentagon below.

This reflects the symmetry ${8 \choose 3} = {8 \choose 5}$, which is a special case of the general identity ${n \choose k} = {n \choose {n - k}}$.
</s>