TITLE: The minimal number of partitions to cover all $k$ tuples
QUESTION [8 upvotes]: The set $N=\{1, 2, \ldots, 2k\}$ can be partitioned into pairs (e.g $(1,2),(3,4),\ldots,(2k-1,2k)$) in $\frac{(2k)!}{k!2^k}$ ways. 
$k$-tuple is subset of size $k$ in $N$. We say that $k$-tuple is covered by partition $\alpha$ if none of pairs of $\alpha$ are in that tuple. For example $(2,4,\ldots,2k)$ tuple is covered by the partition above but $(1,2,\ldots,k)$ is not.
I am interested in the following problem: 

Find the minimal number of partitions of the set $N$ so that all $k$
  tuples are covered.

There are $2k\choose k$ tuples and every partition covers $2^k$ tuples so the answer is $\ge \frac{2k\choose k}{2^k}$. I hope to find an upper bound that is $c\frac{2k\choose k}{2^k}$ where $c$ is constant. Can anybody help on this problem?

REPLY [7 votes]: Denote $N=\frac{2k\choose k}{2^k}$ and choose, say $m=\lceil 10kN\rceil$ independent random partitions (all partitions have equal probability $1/(2k-1)!!$). For any $k$-set $A$, the probability that it is not covered by a single partition equals $1/N$ (indeed, if we denote this probability by $p$, then it does not depend of $A$, and summing up by all choices of $A$ we get $\binom{2k}kp=2^k$, since any partition covers $2^k$ subsets), thus the probability that it is not covered by any of our partitions equals $(1-1/N)^{m}<e^{-10k}$. Summing up by all $2k\choose k$ subsets $A$, we see that the probability that a not covered subset exists does not exceed ${2k\choose k}e^{-10k}<4^ke^{-10k}<1$. Therefore there exists a suitable choice of $m$ partitions.