Problem 14. A set U contains $2n$ elements. We select $k$ subsets of $U$ in such a way that none of them is a subset of another one. What is the maximum possible value of $k$? (Hint: Maximal $k$ is achieved when all subsets have $n$ elements. Indeed, imagine the following process: We start with an empty set and add random elements one by one until we get $U$. At most one selected set can appear during this process. On the other hand, the expected number of selected sets that appear during this process can be computed using the linearity of expectation. Take into account that the probability to come across some set $Z \subset U$ is minimal when $Z$ contains n elements, since all the sets of a given size are equiprobable.)
There is a similar question was already asked before, but I want to prove this following the hint of the author.
With some help from translators and another hint from a Russian site, I have a vague understanding about the problem, but I'm not sure it's correct, and some of my explanation I feel like not clear enough. So my question is:
1) Is it correct?
2) Could you provide a proof based on the hint of the book or refine my proof (the case it's correct)?
Let $K$ is the system of subsets of $U$, such that none of them is a subset of another one.
By adding random elements one by one to an empty set until we get $U$, then we have $2n!$ possible outcomes.
Consider the probability that a set of $K$ appear in the process of selecting elements of $U$, that is, if $M\in K$, all of the first $m=\left|M\right|$ elements of selecting process will be elements of $M$.
Then, for each process, at most only one $M$ will be recognized (if not, more than 1 sets are recognized, there's set of $K$, which $M$ be its subset). Let $\xi_{M}$, $M\in K$ be Bernoulli random variables with, $P\left\{ \xi_{M}=1\right\} =p=\dfrac{1}{2n!}$, the probability of appearance of set M in each process, $P\left\{ \xi_{M}=0\right\} =1-p$.
Let $S_{i}$ is the result of $i$-th selecting process, $S_{i}=\xi_{M_{1}}+\cdots+\xi_{M_{i}}$. Because, at some selection process, no $M$ could be select, so $\mathrm{E}S_{2n!}\leq1$.
Let $t_{m}$ the number of sets with the same cardinality $m$, and $\sum t_{m}=k$: $$\mathrm{E}S_{2n!}= \sum m!\left(2n-m\right)!\cdot t_{m}\cdot p\\ = \sum\binom{2n}{m_{i}}^{-1}t_{m}$$
But, for all $m$, $\dbinom{2n}{n}^{-1}\leq\dbinom{2n}{m_{i}}^{-1}$, and for the case all the set of $K$ have $n$ elements, then each of process of selecting elements we could always select a set of $K$, that is $\mathrm{E}S_{2n!}=1$. We have: $$\sum\binom{2n}{m_{i}}^{-1}t_{m} \leq\sum\binom{2n}{n}^{-1}t_{n}\\ \sum t_{m} \leq\sum t_{n}\\ k \leq \sum t_{n}$$ That is the number $k$ attains maximum when all the set of $K$ have $n$ elements.