I'm interested in the following:


Given a set $S\_{n,k}$ of binary sequences of length $n$ with $k$ many 1-entries, what is the maximal size of a subset $S'\_{k'}\subset S\_{n,k}$ such that for every subset $T\subset S'\_{k'}$ of size $k'$ there is no set of $k'$ many indices on which all elements of $T$ are $1$?


It seems to be quite a natural question to me, so i suppose there are some results. Unfortunately, I didn't find much on finite almost disjoint sets.


Thanks in advance!



Almost disjoint set (finite case)
So recast the problem as: 


$Q(n,k,m,t):$ Given the family of *all* $k$ element subsets of $\{{1,\cdots,n\}}$, how large can a sub-family $\mathcal{F}$ be if we require that no $m$ of them have an intersection of size $t$?


Then your question is $Q(n,k,k',k')$. A very rough bound is that at the absolute largest we could imagine that every subset of of size $t$ is in $m-1$ members of $F$. If $N$ is the size of $F$ then counting in two ways all ways to select a member of $\mathcal{F}$ along with a further subset of size $t$ we have that $$N\binom{k}{t} \le (m-1)\binom{n}{t}$$ so $$N \le \frac{(m-1)\binom{n}{t}}{\binom{k}{t}} \tag{\*}$$


It might be the case that that is not as crude as one might think in the sense that one can prove that there is for fixed $k,m,s$ always a family of size $N$ (rounded down to an integer) provided that $n$ is larger than some $n(k,m,s)$. However that could be a result of the form "there is at least one way to pick the family $\mathcal{F}$ which works" without any real guidance on how to get an actual specimen (nor an assurance that there are any examples with a compact description).


To get equality in $(\*)$ requires the right hand side to be an integer (of course). Then $\mathcal{F}$ is called a $t$-design (with $\lambda=m-1$) and a Steiner system when $m=2$. A very exciting recent paper (link and description [here](http://cameroncounts.wordpress.com/2014/02/06/steiner-systems-exist/)) claims that this integrality condition is sufficient for existence, provided that $n$ is large enough.


