Let $l \leq m $. Let $X$ be a set $|X| = n$ and $Z \subseteq \binom{X}{m}$, so that for every set $L \in \binom{X}{l}$ there is at most one set $B \in Z$ with $L \subseteq B$. How can one prove the following: $$|Z| \leq \frac{\binom{n}{l}}{\binom{m}{l}}\,?$$
I did the following and I'd like to know if it is correct or not.
$$\binom{n}{l} =\frac{n!}{l!(n-l)!} $$ and
$$\binom{m}{l} =\frac{m!}{l!(m-l)!}\,.$$
Dividing both:
$$\frac{\frac{n!}{l!(n-l)!}}{\frac{m!}{l!(m-l)!}} = \frac{n!}{l!(n-l)!} \cdot \frac {l!(m-l)!}{m!}\,. $$
Since $Z \subseteq \binom{X}{m}$, we can say that $$Z \subseteq \frac{X!}{m!(X-m)!}\,,$$
so it follows that
$$\frac{X!}{m!(X-m)!} \leq \frac{n!}{l!(n-l)!} \cdot \frac {l!(m-l)!}{m!}\,. $$
We can divide by $$\frac {l!(m-l)!}{m!}$$
and get
$$\frac{X!}{m!(X-m)!} \cdot \frac {m!}{l!(m-l)!} \leq \frac{n!}{l!(n-l)!}\,. $$
We can cancel $m!$ and get
$$\frac {|X|!}{(|X|-m)! \cdot l!(m-l)!} \leq \frac{n!}{l!(n-l)!}\,.$$
Since $|X| = n$, we can write
$$\frac {n!}{(n-m)! \cdot l!(m-l)!} \leq \frac{n!}{l!(n-l)!}\,,$$
which is true, because the bigger the denominator gets, the smaller the number.
Define $$S:=\Biggl\{(L,B)\in \binom{X}{l}\times Z\,\Big|\,L\subseteq B\Biggr\}.$$ Since for each $\displaystyle L\in \binom{X}{l}$, there exists at most one $B$ in $Z$ for which $L\subseteq B$, we have $$|S|\leq \Biggl|\binom{X}{l}\Biggr|=\binom{n}{l}\,.$$ Now, each $B\in Z$ has $m$ elements, and therefore, there are $\displaystyle\binom{m}{l}$ subsets $\displaystyle L\in\binom{X}{l}$ such that $L\subseteq B$. This shows that $$|S|=\binom{m}{l}\,|Z|\,.$$ The claim follows immediately.