This historical question recalls Pafnuty Chebyshev's estimates for the prime distribution function. In his derivation Chebyshev used the factorial ratio sequence $$ u_n=\frac{(30n)!n!}{(15n)!(10n)!(6n)!}, \qquad n=0,1,2,\dots, $$ which assumes integer values only. The latter fact can be established with the help of $$ \operatorname{ord}_p n! =\biggl\lfloor\frac{n}{p}\biggr\rfloor+\biggl\lfloor\frac{n}{p^2}\biggr\rfloor +\biggl\lfloor\frac{n}{p^3}\biggr\rfloor+\dots $$ and routine verification of $$ \lfloor 30x\rfloor+\lfloor x\rfloor-\lfloor 15x\rfloor-\lfloor 10x\rfloor-\lfloor 6x\rfloor\ge0. $$ Other Chebyshev-like examples of integer-valued factorial sequences are known; the complete list of such $$ u_n=\frac{(a_1n)!\dots(a_rn)!}{(b_1n)!\dots(b_sn)!} $$ in the case $s=r+1$ was recently tabulated in [J.W. Bober, J. London Math. Soc. (2) 79 (2009) 422--444]. A motivation for this classification problem is in relation with a certain approach to Riemann's hypothesis, but I would prefer to refer everybody interested in to Bober's paper (which could be found in the arXiv as well). The proofs of $u_n\in\mathbb Z$ make use of the above formula for $\operatorname{ord}_p n!$

There are three 2-parameter families in Bober's list, namely, $$ \frac{(n+m)!}{n!m!}, \qquad \frac{(2n)!(2m)!}{n!(n+m)!m!}, \qquad\text{and}\qquad \frac{(2n)!m!}{n!(2m)!(n-m)!} \quad (n>m); $$ the first one includes the binomial coefficients, while some properties of the second family are mentioned in this question. For the binomial family, a standard way to establish integrality purely combinatorially amounts to interpreting the factorial ratio as coefficients in the expansion $$ (1+t)^{n+m}=\sum_{k=0}^{n+m}\binom{n+m}{k} t^k, $$ that is, as the number of $m$-element subsets of an $(n+m)$-set. There is lack of similar interpretation for the other two 2-parametric families, although Ira Gessel indicates in [J. Symbolic Computation 14 (1992) 179--194] that the inductive argument together with identity $$ \frac{(2n)!(2(n+p))!}{n!(n+(n+p))!(n+p)!} =\sum_{k=0}^{\lfloor p/2\rfloor}2^{p-2k} \binom{p}{2k} \frac{(2n)!(2k)!}{n!(n+k)!k!} \qquad (p\geq 0) $$ allows one to show that the numbers in question are indeed integers. A slight modification of the formula can be used for showing that the third 2-parametric family is integer valued. In these cases one uses a reduction to binomial sums for which the integrality is already known. But what about the 1-parametric families, like Chebyshev's or, say, $$ \frac{(12n)!n!}{(6n)!(4n)!(3n)!}? $$ Is there any way to establish the integrality without referring to the $p$-order formula?

My own motivation is explained in the joint recent preprint with Ole Warnaar, where we observe a $q$-version of the integrality in a "stronger form".