The maximum number of binary trees that can be formed with three unlabeled nodes is:
For unlabelled tree
T(n) = (2n)! / (n+1)!n!
Number of Labeled Tees = (Number of unlabeled trees) * n!
= [(2n)! / (n+1)!n!] × n!
Can be found with formula... $(2nCn/n+1)$... $n$ being the number of nodes. for the given question... where $n=3$... answer is $5$. Let me also specify here.. that number of Binary Search Trees with $n$ nodes is equal to number of unlabeled Binary trees.
Correct Answer: $B$