**Q1) How many binary search trees possible with 11 distinct key?**

**Ans) **nth Catalan Number $\frac{1}{n+1}\binom{2n}{n}$ = $\frac{1}{11+1}\binom{22}{11}$ = 58786

**Q2)How many binary search trees possible with 11 un-labeled nodes?**

**Ans)** How can we create BST without any key or label?

**Q3)How many binary search trees possible with 11 labeled nodes?**

**Ans) **Binary search trees have keys and they are always labeled. If nodes are different then answer is same as above and if all nodes are same then check answer of 5th question.

https://www.geeksforgeeks.org/how-to-handle-duplicates-in-binary-search-tree/

**Q4) How many binary trees possible with 11 distinct key?**

**Ans)** It is a binary tree and it just have one rule that a parent node can have at max 2 children. (No restrictions like BST's)

$\frac{(2n)!}{(n+1)!}$ = $\frac{(22)!}{12!}$

https://gatecse.in/number-of-binary-trees-possible-with-n-nodes/ (check the 1st answer)

**Q5) How many binary trees possible with 11 un-labeled nodes?**

**Ans) **nth Catalan Number $\large \frac{1}{n+1}\binom{2n}{n}$ = $ \large \frac{1}{11+1}\binom{22}{11}$ = 58786

**Q6) How many binary trees possible with 11 labeled nodes?**

**Ans) **If all labels on nodes are distinct then total numbers of binary trees = $\frac{(22)!}{12!}$

but in case there are k nodes which are having the same label then,

number of binary trees = $\large \frac{(22)!}{12! \ k!}$