k subsets of a set 
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.

For example, if $n=3$ and $k=2$, the set of ordered 2-tuples is: $\{(1,2),(1,3),(2,3),(3,2),(3,1),(2,1)\}$
Here every pair of tuples has at least one common element, hence answer is 6.
Note: $(1,2)$ and $(2,1)$ are different tuples.
 A: As Calvin Lin observed, for every subset of size $k$, there are $k!$ k-tuples. So I'll focus on those subsets first. We'll just have to multiply our answer by $k!$ afterwards.
There are two distinct cases, depending on whether $k$ is greater than $\frac{n}{2}$.
This is because if $k > \frac{n}{2}$, for any given subset of size $k$, it's impossible to construct a second subset of that same size that doesn't contain at least one element of the first subset.
This means the answer for these cases is equal to the total number of subsets, which is $n \choose k$.
Now if $k \le \frac{n}{2}$, I think we need to choose one element to be in all subsets, or it would be possible to construct a second, disjoint subset. So after having picked 1 such element, there are ${n-1}\choose{k-1}$ subsets that can be created containing that element.
So, remembering that we have to multiply by $k!$, the maximum number $m$ of ordered k-tuples is
$$
m(n,k)=\cases {
\begin{align}
{{n}\choose{k}} \cdot & k! & \text{if } k > \tfrac{n}{2}\cr \\
{{n-1}\choose{k-1}} \cdot & k! & \text{if } k \le \tfrac{n}{2}
\end{align}}
$$
A: We are counting the maximum size of a family of ordered $k$-tuples of $[n]:=\{1,2,\dots ,n\}$ such that any two $k$-tuples have a common element, where $1\le k\le n$.
Candidate(1): We suspect that a maximum family is obtained by keeping a element common between all the $k$-tuples (as we do with Erdos-ko-rado when the size the subsets is greater then $\frac{n}{2}$).
Let $1$ be the common element, wlog. We build a $k$-tuple using a $r$-subset of $[n]$ where $1\le r\le k$. Using inclusion-exclusion, we can build a $k$-tuple in $T(r,k)=\sum_{i=0}^{r-1}(-1)^{i}\binom{r}{i}(r-i)^k$ ways. Hence, the maximum size of the family of ordered $k$-tuples is $\sum_{r=1}^{k-1}\binom{n-1}{r}T(r,k)$
Candidate(2) for maximum size when $k>\frac{n}{2}$ is: $\binom{n}{k}k!$. The idea is to select two different $k$-subsets (they will intersect as $k>\frac{n}{2}$) and then generate all the $k$-tuples. Some experimentation with smaller numbers show that as $n$ grows candidate(1) is larger than candidate(2) even when $k>\frac{n}{2}$. Although, for $n=3,k=2$, candidate(2) gives the correct answer.
