How many distinct binary search trees can be created out of 4 distinct keys?
answer - B
number of distinct BSTs = $\frac{^{2n}C_n}{n+1}$ (or) =$\frac{(2n)!}{(n+1)!n!}$ For a given Binary tree structure, there can be only 1 BST. Hence, no. of different BSTs with n nodes will be equal to the no. of different binary tree structures possible for n nodes.
For derivation: http://gatecse.in/number-of-binary-trees-possible-with-n-nodes/
no because distinct keys in BST must satisfy a property. while in a labelled nodes binary tree there is no such restriction
Lakshman Patel RJIT yes try creating distinct binary trees with 3 unlabeled nodes and also try creating BST's with keys {1,2,3} and it is unlabeled Binary tree not unlabeled BST
@Mk Utkarsh I'm little bit confuse
can you explain?
Number of binary trees with n-vertices =2n!/n!(n+1)! =14 so option B is answer
Gatecse