Partitioning $2p$, subject to a divisibility condition. Let $p$ be a prime number and $l$ be the greatest prime number less than $2p$. Moreover, let $m,t_i<p$ be positive integer numbers  such that $\sum_{i=1}^mit_i=2p$. Is it possible that  $1^{t_1}2^{t_2}\ldots m^{t_m}t_1!t_2!\ldots t_m!\mid(2p-l)!$
 A: It is possible. Take $p=61$ and $t_1=2$, $t_{120}=1$. Then $l=113$ and $1^{t_1}\cdot t_1!\cdot 120^{t_{120}}\cdot t_{120}!=240$ divides $(2p-l)!=9!=362880$.
Another example: $p=673$, $t_1=t_{672}=2$. Then $l=1327$ and $1^{t_1}\cdot t_1!\cdot 672^{t_{672}}\cdot t_{672}! = 1806336$ divides $(2p-l)!=19!$.
UPDATE. If all $t_1,\dots,t_m$ are required to be (strictly) positive, then there are likely no required examples. Numerically I tested this for primes $p$ below $10^5$. For large $p$ this can be assessed heuristically as follows.
By Cramer's conjecture, $2p-l$ is asymptotically bounded by $\log(p)^2$. We also have that $m!$ divides $1^{t_1}\cdot 2^{t_2}\cdots m^{t_m}$, which in turn divides $(2p-l)!$, and thus $m\leq 2p-l$. Furthermore, since $\sum_{i=1}^m it_i=2p$, at least one of $t_i$ must be at least $$\frac{2p}{1+2+\dots+m}\geq \frac{4p}{(m+1)^2}\geq \frac{4p}{(2p-l+1)^2}.$$
That is, $t_1!\cdots t_m!\mid (2p-l)!$ would asymptotically imply $\frac{4p}{\log(p)^4}\leq \log(p)^2$, which is not possible.
