Task: Give an example of a partial order in which there are exactly two antichains of size 10, and these antichains do not intersect; and there are exactly 100 chains of size 3 (the size of the chain or antichains is the number of elements in it).
My solution: since we have two anti-chains of size 10 and they do not intersect, this means that if we combine these two anti-chains, they form an anti-chain of size 20. Then, according to Dilworth's theorem, in a given partial order, 20 is the minimum number of chains that is necessary to cover all elements of the partial order. We are also given in the problem that there are 100 chains of size 3. since a chain is a subset of elements of an order in which any two elements are comparable, it turns out that if we take any subset of the elements of the set of elements that form a chain, then this set will also be a chain. It turns out that if we take subsets of those 20 chains that cover the entire order, then we can get 100 chains of size 3. There are a lot of ways to get such 100 chains. Therefore, as an example, we can give such an order that there are exactly 20 chains of 15 elements each.
Please check if my reasoning is correct and I don't quite understand how to give an example of this order. Is it enough to say that one of the variants of this order will be an order of 300 elements, consisting of 20 chains with 15 elements each. Or do I still need to come up with some kind of partial order relation and use it to build that order on the 20 chains mentioned earlier? How are examples of partial orders usually given?
Here is an example:
For $i = 1, \ldots, 10$ define $a_i = \{i\}, \space b_i =\{1, \ldots, 10\} \cup \{i+10\}$
Moreover, $c = \{1, \ldots, 20\}$.
$A = \{a_i: 1 \le i \le 10\}$, $B = \{b_i: 1 \le i \le 10\}$,
$P = A \cup B \cup \{c \}$, partially ordered by inclusion.
Then for all $a \in A, b \in B$ it is $a < b < c$, elements of $A$ are pairwise incomparable, same for elements of $B$. And, in fact, this is all we wanted to achieve.
Hence, $A$ and $B$ are disjoint antichains with 10 elements, each. Moreover, every antichain with more than one element is contained in either $A$ or $B$. Hence, $A, B$ are the only antichains with 10 elements.
Chains in $P$ of size 3 are exactly the sets $\{a,b,c\}$ with $a \in A, b \in B$. Hence, $P$ contains exactly 100 chains of size 3.