Estimate number of subsets 
Let $X_1,\cdots,X_m \in \binom{[n]}{k}, X_i \cup X_j \neq [n], X_i \neq X_j \
 \forall 1 \leq i < j \leq m$.
Show that $k \geq \frac{n}{2} \Rightarrow m \leq \left(1-
 \frac{k}{n}\right)\binom{n}{k}$.

I thought about the Erdős–Ko–Rado theorem. I take the set $\mathcal{X}' = \{X_1^c, \cdots, X_m^c\}, \mathcal{X}' \subseteq \binom{[n]}{n-k}$. $k \geq \frac{n}{2} \Rightarrow (n-k) \leq \frac{n}{2}$, so I can use the theorem for $\mathcal{X}'$ if I can show that $X_i^c \cap X_j^c \neq \emptyset \ \forall 1 \leq i < j \leq m$. The theorem says that $|\mathcal{X}'| \leq \binom{n-1}{(n-k)-1} = \binom{n-1}{n-k-1}$.
I do have two problems with that approach:


*

*How do I show that $X_i^c \cap X_j^c \neq \emptyset \ \forall 1 \leq i < j \leq m$ applies?

*How do I show that $\binom{n-1}{n-k-1} = \left(1-\frac{k}{n}\right)\binom{n}{k}$? I tried to rearrange it for some time, but still couldn't figure it out.


(1) for $1\leq i<j\leq m$,
$$\eqalign{X_i\cup X_j\ne [n]&\iff (X_i\cup X_j)^c\ne [n]^c\cr
&\iff  X_i^c\cap X_j^c\ne \emptyset\cr}$$
(2) $$\binom{n-1}{n-1-k}=\frac{(n-1)!}{k!(n-k-1)!}
=\frac{n-k}{n}\cdot\frac{n!}{k!(n-k)!}=\left(1-\frac{k}{n}\right)\binom{n}{k}.$$
