semi-standard fillings of $\lambda$ by real numbers in $(0,1]$ where semi-standard mean weakly increasing in rows and strictly increasing in columns. Clearly, $\mathcal{Y}_{\lambda}$ is a (partially open) convex polytope in $\mathbb{R}^\lambda$. A semi-standard tableau with largest entry less than $t$ is just an element of $\mathbb{Z}^{\lambda} \cap t \mathcal{Y}_{\lambda} $, so that $$E(\mathcal{Y}_\lambda,t) = s_{\lambda}(1^t),$$ the Schur function $s_{\lambda}$ where $t$ variables are set to $1$ and the others to $0$. Let $SYT(\lambda)$ denote the set of standard tableaux of shape $\lambda$, and recall that a descent of a standard tableau is an entry $i$ such that the entry $i+1$ is in an upper row. Let $\operatorname{des}(T)$ denote the number of descents of a standard tableau, then we have:

$$\begin{aligned}
 \label{sytdes}
  E^*( \mathcal{Y}_{\lambda} , z ) = (1-z)^{n+1} \sum_{ t\geq 0 } s_{\lambda}(1^t) z^t = \sum_{ T\in SYT(\lambda) } z^{\operatorname{des}(T)+1}.\end{aligned}$$

Let us sketch the proof. The reading word $w(T)$ of a semi-standard tableau $T\in\mathcal{Y}_\lambda$ is defined by ordering its entries row by row, from left to right and from top to bottom. Then, the standardization $\operatorname{std}(T)$ is defined to be the unique standard tableau $U$ of the same shape such that $\operatorname{std}(w(T)) = w(U) $. The set $\mathcal{Y}_{\lambda}$ is partitioned 