How large should my family of subsets be before I am guaranteed an element meeting some requirements? Let $X$ be a set, and let $a_1,\ldots, a_k$ be some distinguished elements of $X$. Let $F$ be a family of subsets of $X$, where the maximum cardinality of any set in $F$ is at most $m < |X|$. Suppose also that $k < X-m$. What is the smallest size of $|F|$ such that I know for sure that there is an $m$-set in $F$ that does not contain any of the $a_1,\ldots, a_k$?
My initial thought was that if 
$$  |F| > \sum\limits_{i=0}^{m-1} \binom{|X|-k}{i} + \sum\limits_{A\subset \{a_1\ldots a_k\}} \sum\limits_{i=0}^{m-|A|}\binom{|X|-|A|}{i},$$
then this should suffice, because then $|F|$ exceeds the total possible number of sets not larger than $m-1$ outside of $a_1,\ldots, a_k$, plus the total number of sets of size no more than $m$ that contain at least one of $a_1,\ldots, a_k$. So by the pigeonhole principle this bound should suffice.
Is this bound correct? And is there any upper bound that is perhaps slightly more manageable than the one presented.
Thanks in advance!
P.S. Sorry for the unclear title, I did not know how else to phrase this problem briefly.
 A: (Updated 1:30pm EST when I noticed the $|Y|=m$ requirement...)
I am not sure I understand your bound, but here's how I would do it.  (I'm not sure if this is equivalent or different.)
Let $n= |X|$ and $A = \{a_1, \dots, a_k\}$.  Also, we'll say $F$ is feasible if $F$ contains an $m$-subset that does not intersect $A$, i.e. $\exists Y \in F$ s.t. $Y \subset X, |Y| = m, Y \cap A = \emptyset$.
Since any subset of size $< m$ cannot make $F$ feasible, the maximum infeasible $F$ will contain all such subsets.  There are $\sum_{i = 0}^{m-1} {n \choose i}$ of them.  (Note that many of these will not intersect $A$, but they do not make $F$ feasible because the requirement includes $|Y|=m$.)
Further, the maximum infeasible $F$ contains all $m$-subsets which intersect $A$.


*

*Total number of $m$-subsets $= {n \choose m}$

*Total number of $m$-subsets which do not intersect $A = {n - k \choose m}$

*So, total number of $m$-subsets which intersect $A = {n \choose m} - {n - k \choose m}$
Therefore the maximum infeasible $F$ has $\sum_{i = 0}^{m-1} {n \choose i} + {n \choose m} - {n - k \choose m}$ subsets.  If $|F|$ is bigger than this number, then $F$ is guaranteed feasible.
A: Before satisfying your condition, $F$ could contain all sets with cardinality $\leqslant m-1$ and also all sets with cardinality $m$ that intersect $A=\{a_1,\dots,a_k\}$.
Let $n = |X|$.
For the first kind, there would be $\sum_{j = 0}^{m-1} \binom n j$ sets.
For the second kind, there are a total $\binom n m$ sets with cardinality $m$, of which $\binom{n-k}m$ do not intersect $A$.
Therefore, there are $\binom n m - \binom {n-k}m$ sets with cardinality $m$ that intersect $A$.
The smallest size that guarantees $F$ would satisfy your condition is hence
$$1 + \binom n m - \binom {n-k}m + \sum_{j = 0}^{m-1} \binom n j
= 1 + \sum_{j=0}^m\binom n j - \binom{n-k}m.$$
