6 permutations of $[1,2,\cdots,n]$ so for all distinct $a,b,c\in \{1,2,\cdots,k\}$, only one permutation has $a$ before $b$ and $b$ before $c$. Consider an integer $k \ge 3$. Prove that there exists a $k$ such that there are no $6$ permutations of $1,2,3,\cdots, k$ that satisfy the following condition:
For all distinct numbers $a,b,c\in \{1,2,\cdots,k\}$, there is exactly one permutation out of the six that has the number $a$ before $b$ and $b$ before $c$.

So far, I have found that in the six permutations, the ordering of $a,b,c$ have to be distinct. This is because if there are two decks have the same ordering, say, $b, c, a$, then the triplet $b,c,a$ won't satisfy the condition. I think I have to find some sort of contradiction from this, but I'm not sure how.
Thanks in advance!!
 A: Build on your observation
You observed that the ordering of $a,b,c$ must be distinct. Since there are only $6$ orderings of triplets, each of these appear exactly once in the $6$ selected permutations of $\{1,...,k\}$.
Helper Classification
Choose two arbitrary distinct numbers, say $m$ and $n$. For each number $a_{i}\neq m,n$, we assign a six digit number $h_{i1}h_{i2}h_{i3}h_{i4}h_{i5}h_{i6}$ with the following rule:
$$
h_{ij}=
\begin{cases}
1, \text{ if } a_{i} \text{ is positioned before m & n in permutation j} \\
2, \text{ if } a_{i} \text{ is positioned between m & n in permutation j} 
\\
3, \text{ if } a_{i} \text{ is positioned after m & n in permutation j} 
\end{cases}
$$
There are only $\frac{6!}{2!2!2!}=90$ distinct six digit numbers like this. Pigeon-Hole-Principle, for $k\geq 93$ there is always a pair of number with identical six digit number.
Final Step and Conclusion
If $a$ and $b$ has identical six digit number, both of them appear before $m$ in exactly $3$ permutations. However, we only have two allowed orderings: $a,b,m$ or $b,a,m$. Therefore, if $k\geq 93$, there are no six permutations that satisfy your condition.
