VMO

a) The answer is $|T|=2n(2n-1)(2n-2)$.

b) Consider the set $$A=\{(i,j):1\le i,j\le n,i\ne j\}\cup \{(i,j):n+1\le i,j\le 2n,i\ne j\}.$$ Clearly, $|A|=2\cdot A_n^2=2n(n-1)$. Furthermore, for all $(x,y,z)\in T$ there always exist two in three numbers $x,y,z$ belong to the same set $\{1,2,\dots ,n\}$ or $\{n+1,n+2,\dots ,2n\}$. Without loss of generality, assume that $x,y$ belong to the same set then $(x,y)\in A$, which implies $\{(x,y),(x,z),(y,z)\}\cap A\ne \varnothing$.

c) Consider the graph $G_1=(V,E_1)$, $V$ is the set of vertices that represents for $1,2,\dots ,2n$ and $u,v$ are connected when a set $A$ which is connected with $T$ does not have the ordered pair $(u,v)$. From the given condition, at least one of three numbers $\{x,y,z\}$ belongs to $A$, otherwise there exists three segments which connect $(x,y)$; $(y,z)$; $(z,x)$. Without loss of generality, suppose that $x>y>z$ then $G_1$ contains $(x,y)$, $(x,z)$, $(y,z)$, which means when we consider the triple $(x,y,z)$ then none of these ordered pairs belong to $T$, contradiction. Therefore, $G_1$ does not contain a triangle. By Mantel - Turan theorem, the number of edges in $G$ cannot exceed $\frac{(2n{)}^2}{4}=n^2$. Hence, the number of edges in the complementary graph is at least $\left(\genfrac{}{}{0ex}{}{2n}{2}\right)-n^2=n(n-1)$. Similarly, we can show that there are at least $n(n-1)$ pairs $(u,v)$ $(u<v)$ in $A$. Hence, $T$ has at least $2n(n-1)$ ordered pairs. $\square$