Minimum number of subsets of combinations so all combinations in a subset differ by one Consider all ${n \choose k}$ subsets of size $k$ from a larger set of size $n$.  I would like to split the set of subsets into parts so that in each part, the intersection of all the subsets is at least of size $k-1$.  

What is the minimum number of parts that you are guaranteed to be able
  to split the set of subsets into?

The maximum number of subsets in a part is $n$ so this at least gives us a lower bound of $\frac{{n \choose k}}{n}$ for the number of parts. But this is not tight.
When $k=1$ the minimum number of parts is $1$. When $k=2$ the minimum number of parts is $n-1$. I don't know how to solve the problem for $k=3$.
Edit
Smylic points out you can slightly raise the lower bound as each  part has at most $n  - (k-1)$ subsets. So this makes a new lower bound of $\frac{{n \choose k}}{n  - (k-1)}$.
 A: This is not a complete answer, just an upper bound.
A trivial upper bound is $\binom{n}{k - 1}$, because each part has its specific subset of size $k - 1$ that is subset of each element of part (even is a part has only one subset of size $k$ it can be assigned some subset of size $k - 1$).
I'm going to show that $\binom{n - 1}{k - 1}$ for $k \ge 1$ is a better upper bound (and $k = 0$ is a trivial case with the only part of one empty subset). Let $f(n, k)$ be the desired minimum number of parts. Then $f(n, 1) = 1$ and $f(n, n) = 1$. Suppose $f(m, k) \le \binom{m - 1}{k - 1}$ for all $m < n$ and $1 \le k \le m$. We have $f(n, k) \le f(n - 1, k - 1) + f(n - 1, k)$ since we can firstly divide all subsets with element $n$ into $f(n - 1, k - 1)$ parts and then divide all subsets without element $n$ into $f(n - 1, k)$ parts. So
$$f(n, k) \le f(n - 1, k - 1) + f(n - 1, k) \le \binom{n - 2}{k - 2} + \binom{n - 2}{k - 1} = \binom{n - 1}{k - 1}$$
for $2 \le k \le n$ and $f(n, 1) = 1 \le 1 = \binom{n - 1}{0}$. Thus we have proven an induction step and the basis is $n = 2$.
This upper bound is less than $k$ times greater than the lower bound achieved before:
$$\frac{\binom{n - 1}{k - 1}}{\frac{\binom{n}{k}}{n - (k - 1)}} = \frac{(n - k + 1)}{n}k.$$
However $f(6, 3) = 6$ shows that both of them are not tight.
A: I have a solution for $k=3$; when $k>3$, the question is a well-known open problem.
When $k=3$, instead of dividing the $\binom{n}{3}$ subsets into parts, it's equivalent to choose some number of pairs $\{x,y\} \subset \{1,2,\dots,n\}$ to serve as their intersections. (In most cases, a part will have intersection of size exactly $2$; if a part consists of only one subset $\{x,y,z\}$, we may arbitrarily choose any of $\{x,y\}$, $\{x,z\}$, or $\{y,z\}$ to be its assigned pair.) Conversely, if we specify some pairs such that every $3$-subset contains at least one pair, we have specified the partition, up to some flexibility when a $3$-subset contains more than one pair, and can fit in more than one part.
We can think of choosing these pairs as choosing a graph $G$ on vertex set $\{1,2, \dots, n\}$. The condition that every $3$-subset is assigned a part is now saying that for any $\{x,y,z\}$, one of $xy$, $xz$, or $yz$ must be an edge of $G$. Equivalently, the complement $G^c$ is a triangle-free graph.
By the $r=2$ case of Turán's theorem, a triangle-free graph on $n$ vertices can have at most $\left\lfloor\frac{n^2}{4}\right\rfloor$ edges, and this is tight: the bipartite graph $K_{\lfloor n/2\rfloor, \lceil n/2\rceil}$ has that many edges and is triangle-free. Therefore $G$ has at least $\binom{n}{2}-\left\lfloor\frac{n^2}{4}\right\rfloor$ edges, and any partition of the $3$-subsets of $\{1,2,\dots,n\}$ of the kind we're looking for has at least $$f(n,3) = \binom{n}{2} - \left\lfloor\frac{n^2}{4}\right\rfloor = \left\lfloor\frac{(n-1)^2}{4}\right\rfloor$$ parts.
Similarly, for $k>3$, we can interpret the problem as a hypergraph Turán problem. Instead of the graph $G$, we have a $(k-1)$-uniform hypergraph $H$, whose complement $H^c$ contains no $k$-clique $K_k^{k-1}$. The Turán number $\DeclareMathOperator{\ex}{ex}\ex(n,K_k^{k-1})$ is the maximum number of edges $H^c$ can have. The answer $f(n,k)$ we're looking for is the minimum number of edges $H$ can have, so $$f(n,k) = \binom{n}{k-1} - \ex(n,K_k^{k-1}).$$ This value is also often denoted $f(n,k) = T(n, k, k-1)$.
It's known that as $n \to \infty$, $$T(n, k, k-1) \sim t(k,k-1) \binom{n}{k-1}$$ for some constant $0<t(k,k-1)<1$ independent of $n$. As far as I know, $$(1+o(1))\frac1{k} \le t(k,k-1) \le (1+o(1))\frac{\log k}{2k}$$ are asymptotically (as $k \to \infty$) the best bounds known, which I got from this survey, but there may have been new, better results in the past five-six years. (The value of $t(4,3)$ is known to be somewhere between $0.438334$ and $\frac49$, which is also elaborated on in the survey.)
