Divisibilty of binomial coefficients with equal base Is there any criteria for $k,m,n \in N$ such that
${n \choose k}$ diviides ${m \choose k}$.
 A: I don't know of any criteria.  I have a suggestion that might point you to work of Erdos, Selfridge, and others on products of consecutive integers, but I have no specific references for you.
Let c(m,k) be the numerator of a common form of the binomial coefficient, namely $\prod^{k-1}_{i=0} m-i$. Your condition implies c(n,k) divides c(m,k) as well as that n choose k divides a product of a subset of k consecutive integers.  Now any two integers greater than k that are less than k apart are "mostly coprime", meaning that any factors they have in common are smaller than k.  If you have a somewhat smooth number like n choose k being a factor of the product of a subset of mostly coprime numbers, it seems much more likely that it is a factor of just one of the numbers of the subset than not.  Otherwise you have to break up n choose k into two or more mostly coprime factors f_i and arrange all the f_i to have multiples that fall into the same small interval.  While this division may be possible, I suspect it will be hard to arrange for three or more factors of n choose k.  I would start by characterizing those m,n,k,l in which n choose k divides m(m-l) with l less than k.  You may find that a criterion will be that n choose k has such a mostly coprime decomposition that "projects" into a small interval.
Gerhard "Takes Smooth With The Rough" Paseman, 2016.10.05.
