This problem appeared on a qualifying exam

Let $$n\leq 2k$$ and $$A_1,\dots,A_m$$ be subsets of size $$k$$ of $$A=\{1,\dots,n\}$$ with the following property: $$A_i \cup A_j \not = A$$ for all $$i,j$$. Show that $$m \leq (1-\frac{k}{n}) \binom{n}{k}$$.

Attempt: Firstly, we can observe that with some algebra $$(1-\frac{k}{n}) \binom{n}{k}=\binom{n-1}{k}$$. Now, I am thinking that we can use a similar double counting argument as in Katona's proof of EKR. We see that if a given cyclic ordering of $$\{1,\dots,n\}$$ contains some $$A_i = (a_1,\dots,a_k)$$ in the cyclic ordering, then for every $$a_i, a_{i+1}$$ the family $$\{A_1,\dots,A_m\}$$ may admit at most one of the $$k$$ length sets separating $$a_i$$ from $$a_{i+1}$$ because $$2k \geq n$$ and no two sets may union to $$\{1,\dots,n \}$$. So now consider the set of all pairs $$P=(A_i,\sigma)$$, where $$A_i \in \{A_1,\dots,A_m\}$$ and $$\sigma$$ is some interval order. We have by our above observation that if $$\mathcal{P}$$ is the set of all pairs, then $$|\mathcal{P}| \leq k(n-1)!$$. Also, $$|\mathcal{P}| = k! (n-k)!$$ because for every set $$A_i$$ we can arrange it into a cyclic order by permuting its elements then permuting the remaining $$(n-k)$$ elements.

My problem is that this does not give the desired upperbound as this is exactly Katona's proof. Is there an error in this work? Thank you.

Hint: Let $$B_i = A - A_i$$.
Hence conclude that $$m \leq { n-1 \choose n-k - 1 }$$ as desired.