For example, if number of nodes are 3 then 5 unlabelled binary trees[2nCn/(n+1)] are possible with 3 nodes. So, for each unlabelled binary tree there are 6 possible labelled binary trees[(i.e.,n!,(3!=3*2*1)] . Suppose that 3 labelled nodes are 1,2,3 and assume that the inorder traversal for this labelled binary tree is 123, then among those 6 possible binary trees you can get only one binary search tree as inorder 123.