Sorry for the vague title but I'm not familiar with the area and don't know what the right termilogy might be.
The problem I'd like to solve is: Given a collection $S$ of all the size $k$-subsets of the set $N=[1,\cdots,n]$, where $k\leq n$, find the partitions $S_i$ of $\mathcal{S}$ such that each $S_i$ covers $N$, where $S_i \bigcap S_j = \varnothing$ if $i\neq j$, and $\bigcup S_i=S$.
There are $\binom{n}{k}$ members in $S$.
An example is: $N=\{1,2,\cdots,4\}$, $k=2$, so $S=\{ \{1,2\},\{1,3\},\{1,4\},\{2,3\},\{2,4\},\{3,4\} \}$, and a partition would be
$S_1=\{ \{1,2\},\{3,4\} \}$
$S_2=\{ \{1,3\},\{3,4\} \}$
$S_3=\{ \{1,4\},\{2,3\} \}$
I see that the $k$-disjoint set cover problem is NP-complete, but this problem can be seen as a special case where the set $S$ is given as above.
I appreciate it if there are any suggestions on possible directions to solve the problem. Thank you.
The algorithm is simple: to partition $N$ into subsets of size $k$ pick an arbitrary element $x \in N$. Then for each of the $\binom{|N| - 1}{k-1}$ subsets $s$ of size $k$ which contain $x$, recur on $N \setminus s$.
This also gives a simple formula for the number of ways of partitioning $kd$ elements into subsets of size $k$: $$\prod_{i=1}^d \binom{ki-1}{k-1} = \frac{1}{(k-1)!^d} \prod_{i=1}^d\frac{(ki-1)!}{(k(i-1))!} = \frac{(kd-1)!}{(k-1)!^d \prod_{i=1}^{d-1} ki} = \frac{k(kd-1)!}{k!^d (d-1)!} $$
By Stirling's formula $n! \approx \sqrt{2\pi n} \left(\frac n e\right)^n$, this is approximately $$\frac{e^d (kd-1)^{kd-1/2}}{(\sqrt{2\pi})^{d} k^{kd+d/2-1} (d-1)^{d-1/2}} \approx \left(\frac{e d^{(k-1)}}{\sqrt{2\pi k}}\right)^d$$ which is most certainly not polynomial in $d$, so generating all of the solutions is not in NP.