Suppose I am given a set of $n$ pairs of items (so I have $2n$ items in the set). I wish to partition the set into 2 disjoint sets such that at least one pair of items has a member in each set. I want to know how many 2-partitions there are of the $2n$ items where the smallest set has $k$ elements.

For example, if I have two pairs of items $\{a_1,a_2,b_1,b_2\}$ I can partition them into the following subsets:

  • $\{a_1,a_2,b_1\}$$\{b_2\}$;
  • $\{a_1,a_2,b_2\}$$\{b_1\}$;
  • $\{a_1,b_1,b_2\}$$\{a_2\}$;
  • $\{a_2,b_1,b_2\}$$\{a_1\}$;
  • $\{a_1,b_1\}$$\{a_2,b_2\}$;
  • $\{a_1,b_2\}$$\{a_2,b_1\}$;

So there are 4 partitions with the smallest set of size 1, and 2 partitions with the smallest set of size 2.

I have exhaustively computed the numbers for 1, 2, and 3 pairs, but I'm looking for a general formula. I know if $k=1$ there are $2n$ partitions, but I haven't found a closed form for a general solution, or even a recursive formula.

edit: I can show that if $k=2$ there are $4n-4$ partitions.


1 Answer 1


First count the total number of partitions where the smallest set has $k$ elements, and then count the bad partitions, in which all pairs are together. Subtract bad from total.

The situation is a little different for $k=n$ than for $k\lt n$. This is because in the case $k=n$ there are $\frac{1}{2}\binom{2n}{n}$ partitions, and for $k\lt n$ there are $\binom{2n}{k}$.

Let us deal with $k\lt n$. If $k$ is odd, there is no bad partition. At least one couple must be separated, else our $k$-set would be made up of couples, so would have an even number of elements.

Now we deal with the case $k$ even, say $k=2l$. To make a bad partition, we choose $l$ couples. This can be done in $\binom{n}{l}$ ways.

A similar calculation takes care of the case $k=n$.

  • $\begingroup$ Thanks, this was perfect. Just to clarify, for the case where $k=n$, the same rules with whether k is even or odd for $k<n$ apply. If $k$ is even, then there are an odd number of pairs, and any manner of splitting them into two groups will have at least one pair with an element on both sides. $\endgroup$ Jul 18, 2013 at 11:43
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    $\begingroup$ Yes, analysis is the same, no bads if $n$ odd. If $n$ even, say $2l$, then $\frac{1}{2}\binom{2l}{l}$. So only different thing is the $1/2$ in front of both total and bads. $\endgroup$ Jul 18, 2013 at 11:52

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