Let $P$ be a finite poset and let $\,\mathcal{C}=(C\_1,\ldots,C\_\ell)\,$ be its decomposition into chains. We can define
$$
f(\mathcal{C}) = |C\_1|! \, \cdots \,|C\_\ell|!
$$
and ask for what $\mathcal{C}$ we have $f(\mathcal{C})$ maximal. Roughly speaking, the smaller is $\ell$ and the less evenly distributed are $|C\_i|$, the better. Specifically, I am interested in the following. 


**Question:** *Does $f(\mathcal{C})$ maximize for the number $\ell$ of chains equal to the size of the maximal antichain in $P$?*


**Motivation:** If true, this would be an unusual extension of [Dilworth's theorem](https://en.wikipedia.org/wiki/Dilworth%27s_theorem). I am really interested in bounds on the number $e(P)$ of linear extensions, and it is easy to see that $e(P)\le |P|!/f(\mathcal{C})$ for all chain decompositions $\mathcal{C}$. The connection to entopy is also standard: take $\log e(P)$ in the last equation and think of $|C\_i|/|P|$ as a probability distribution. 


**Refs disclaimer:** There is a large number of papers bounding $e(P)$ in terms of the entropy, see e.g. [here](http://www.cdam.lse.ac.uk/Reports/Files/cdam-2003-19.pdf), [here](http://research.microsoft.com/en-us/um/redmond/groups/theory/jehkim/papers/entropy.pdf) and [here](https://arxiv.org/abs/0911.0086). They are not completely unrelated, but do not seem concerned with this type of questions. 



Entropy of chain decompositions of posets
The answer is **no**. Take a chain $a\_1>\dots>a\_{100}$ and add two elements $b>a\_{51}$ and $c<a\_{50}$ (with no relations not implied by these ones). The maximum of $f(\mathcal C)$ is achieved on the decomposition $\{a\_1,\dots,a\_{100}\}\sqcup\{b\}\sqcup\{c\}$, while the maximal antichain has cardinality 2 (due to the decomposition into two chains $\{a\_1,\dots,a\_{50},c\}\sqcup \{b,a\_{51},\dots,a\_{100}\}$).


