Define a (set-valued) function $ n \to \{\dfrac{n!}{k!(n-k)!k}:k=1,2,...n-1,n\}$. Now let $f$ be a function that counts the number of natural numbers in the set $\{\dfrac{n!}{k!(n-k)!k}:k=1,2,...n-1,n\}$.
A question is: Is $f(n)=1$ for only a finite number of values of $n$?
Here, we study, for each natural number $n$, a set $\{\dfrac{n!}{k!(n-k)!k}:k=1,2,...n-1,n\}$ and seek to determine is there a finite number of natural numbers $n$ for which only the first term is a natural number.
Also, investigations of values of function $f$ are really welcomed, it would be nice to see, is, for example, $f(n)$ in OEIS, for some relatively small number of first values of $f$?
I expect a finite number of natural numbers for which $f(n)=1$ because it seems to me that, although there will be much cancellations in calculation of a number $\dfrac{n!}{k!(n-k)!}$, there should be enough richness in a structure of $\dfrac{n!}{k!(n-k)!}$ that allows $\dfrac{n!}{k!(n-k)!k}$ to be a natural number for some $k \in \{1,2,...,n\} \setminus \{1\}$