We are given a set of \(n\) distinct elements and an unlabeled binary tree…

Final answer: There is exactly one way to populate the given unlabeled binary tree so that it becomes a binary search tree.

Key idea: Inorder traversal of a binary search tree visits the keys in increasing order.

• Sort the given n distinct elements.

• Traverse the fixed tree shape in inorder and assign the sorted elements to the nodes in that order (smallest to first inorder node, next to second, and so on).

• Because the BST property requires the inorder sequence to be sorted, this construction yields a valid BST and no other assignment can satisfy the inorder-sorted requirement. Hence the assignment is unique.

Notes on other counts:

• The value 0 is incorrect because at least one valid labeling exists (the inorder-sorted assignment).

• The value n! counts all possible bijections of elements to node positions without the BST constraint, so it overcounts.

• The Catalan number (1/(n+1) * C(2n,n)) counts the number of different unlabeled binary tree shapes with n nodes, not the number of labelings of a fixed shape into a BST.