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?

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?