Note: The position begins with 1.

Given the preorder traversal of a Binary Search Tree (BST):

1. Reconstruct the BST:

   • Root: $50$

   • Left subtree of 50 (elements $< 50$): $[25, 13, 40, 30, 47]$

     • Root is $25$.

     • Left of $25$ ($< 25$): $13$

     • Right of $25$ ($> 25$): $[40, 30, 47]$

       • Root is $40$, left child is $30$, right child is $47$.

   • Right subtree of 50 (elements $> 50$): $[75, 60, 70, 80, 77]$

     • Root is $75$.

     • Left of $75$ ($< 75$): $[60, 70]$

       • Root is $60$, right child is $70$.

     • Right of $75$ ($> 75$): $[80, 77]$

       • Root is $80$, left child is $77$.

2. Find the Postorder Traversal (Left, Right, Root):

   • Postorder of left subtree of $50$:

     • Left of $25$: $13$

     • Right of $25$: $30, 47, 40$

     • Root: $25$

     • Sequence: $13, 30, 47, 40, 25$

   • Postorder of right subtree of $50$:

     • Left of $75$: $70, 60$

     • Right of $75$: $77, 80$

     • Root: $75$

     • Sequence: $70, 60, 77, 80, 75$

   • Root: $50$

3. Full Postorder Traversal: $$\begin{array}{ccccccccccc} 1 & 2 & 3 & 4 & 5 & 6 & 7 & 8 & 9 & 10 & 11 \\ 13 & 30 & 47 & 40 & 25 & 70 & 60 & 77 & 80 & 75 & 50 \end{array}$$

Element $60$ is at position 7 (1-indexed).