# Number of Unordered Binary Tree and Its Recurrence Relation

Let's define unordered binary tree as a tree where each internal node has two children and the relative order of the subtrees of a node is not important. Then show:

$$b_n = {1\over 2}\sum_{i=1}^{n-1}\binom{n}{i}b_ib_{n-i}$$

When we construct cobinatorial argument, my intuition says summation(represented in $\sum$) corresponds to the disjoint union of countings and multiplication is a combinational countings between two independent situation(represented in $b_{i}b_{n-i}$).

If I slightly change the given recurrence relation as :

$$2b_n = \sum_{i=1}^{n-1}\binom{n}{i}b_ib_{n-i}$$

I could think of my self this counting cares about the order of the spanning of the first node so counts twice for each case of order.

However, still what is unclear is that what $\sum_{i=1}^{n-1}$ refer to. It represents certain disjoint dissection on given combinatorial situation,but I still fall in labyrinth.

Any hint to proceed?

The gist of the relation is that you're constructing trees of size $n$ by splitting $n$ nodes into two non-empty groups. Each group is made into a tree and their roots are connected by a new edge. Because the structures of the two trees are independent, the number of such tree pairs for some size $i$ is obtained by multiplying $b_i$, $b_{n-i}$, and the number of ways of splitting the nodes into a group of size $i$ and a group of size $n - i$.
The summation, then, is over $i$, or the size of one of the sub-trees. One tree is made of $i$ nodes, and the other of $n - i$ nodes. All non-empty sub-tree sizes that sum to $n$ are possible, so we need to add them all up. You are correct in that we count each tree twice since the order of the children doesn't matter. A tree made of two children of sizes $k$ and $n - k$ is the same as a tree made of two children of sizes $n - k$ and $k$.