find the maximum number of ordered k-tuples I would like to know the answer for the following.
Given a set of n elements, find the maximum number of ordered k-tuples possible such that every pair of k-tuples has at least one element in common but has no element is common in every k-tuple.
 A: Without loss of generality we let the set be $0,1,2\dots n-1$.
If $n<k$ the answer is $0$ since we cannot even form a $k$-tuple that does not repeat elements. If $n\geq k$ then the answer is $(n-1)\cdot(n-2)\dots (n-k+1)$.
It is clear this can be reached by taking all the $k$-tuples that start with $0$.
Can we have more? No, for starters notice that there are $n\cdot(n-1)\cdot(n-2)\dots (n-k+1)$ $k$-tuples that don't repeat elements in total. So we are proposing the maximum is $\frac{1}{n}$ of all such $k$-tuples.
To see this is the case consider for each $k$-tuple the set of size $n$ consisting of all the $k$-tuples obtained by adding a number between $0$ and $n-1$ to each of the elements of the $k$-tuples and then reducing $\bmod n$.
So for example if $n=5$ and $k=3$ the set corresponding to $(0,2,3)$ is the following:
$\{(0,2,3),(1,3,4),(2,4,0),(3,0,1),(4,1,2)\}$.
We have partitioned the $k$-tuples that don't repeat elements into sets of size $n$. Since there are $n(n-1)(n-2)\dots (n-k+1)$ such $k$-tuples in total and the sets are of size $n$ each there must be $(n-1)(n-2)\dots (n-k+1)$ such sets.
Notice $k$-tuples of the same set do not have elements in common, therefore we can pick at most one from each set. This proves we can have at most $(n-1)(n-2)\dots (n-k+1)$ tuples with non-repeated elements if we want any two of them to have an element in common.
