Define $\binom{a_1,...,a_n}{b_1,...,b_m}=\Pi_i(a_i!)/\Pi_j(b_j!)$. (WLOG nullpad to have $m=n$ and sort descendingly.) It's natural to ask when the value is an integer. Since I am working with "majorization" currently...but alas, $\binom{3,1}{2,2}=3/2$ and $\binom{6,1,1}{5,3,0}=1$ are counterexamples for both directions :/ Also the last example (inverted) shows that not even $a_1>b_1$ is necessary.
Do you know conditions (beside the obvious one that a multinomial coefficient is always an integer)? Any research on this area?
Mordell, L. J. (1959),
Integer Quotients of Products of Factorials.
Journal of the London Mathematical Society, s134: 134–138.
doi:10.1112/jlms/s134.2.134
Errata:
On a Problem of Hardy and Littlewood Volume s134, Issue 4, 485,
Errata:
Integer Quotients of Products of Factorials Volume s134, Issue 4, 485,

$\begingroup$ I can only check 1 and I check the one that proves I was too lazy for the research :) Sorry, Dirk. $\endgroup$ – Hauke Reddmann Dec 13 '17 at 17:01
Let $\nu_p$ be the $p$adic valuation on the integers for a prime $p$, i.e. for an integer $x$ we have $\nu_p(x) = a \in \mathbb{N}_0$ if $p^a \mid x$ and $p^{a+1} \not\mid x$.
Then the following theorem is wellknown from elementary number theory:
For $x \in \mathbb{N}$ we have $$\nu_p(x!) = \sum_{i=1}^{\infty} \left\lfloor \frac{x}{p^i}\right\rfloor.$$
Note that this is a finite sum, as we will only get zeros once $i$ gets too big.
Now your number is an integer if and only if the $p$adic valuation of the first product is bigger or equal then the one of the second product for all primes $p$; or in symbols:
$$\sum_{k=1}^n \sum_{i=1}^{\infty} \left\lfloor \frac{a_k}{p^i}\right\rfloor \geq \sum_{\ell=1}^m \sum_{i=1}^{\infty} \left\lfloor \frac{b_\ell}{p^i}\right\rfloor\,\,\,\, \forall \,\, p \in \mathbb{P}.$$