It will be **2n!/(n+1)!n!**

See the following code

```
int countTrees(int numKeys) {
if (numKeys <=1) {
return(1);
}
else {
// there will be one value at the root, with whatever remains on the left and right each forming their own subtrees.
// Iterate through all the values that could be the root...
int sum = 0;
int left, right, root;
for (root=1; root<=numKeys; root++) {
left = countTrees(root - 1);
right = countTrees(numKeys - root);
// number of possible trees with this root == left*right
sum += left*right;
}
return(sum);
}
}
```