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Total number of possible Binary Search Trees and Binary Trees with n keys | A function to find factorial of a given number.
Calculate value of [ 1 * ( 2 ) * -- - * ( n - k + 1 ) ] / [ k * ( k - 1 ) * -- - * 1 ].
Since C ( n , k ) = C ( n , n - k ).
Calculate value of [ n * ( n - 1 ) * -- - * ( n - k + 1 ) ] / [ k * ( k - 1 ) * -- - * 1 ].
A Binomial coefficient based function to find nth catalan number in O ( n ) time.
Calculate value of 2 nCn.
return 2 nCn / ( n + 1 ).
A function to count number of BST with n nodes using catalan.
find nth catalan number.
return nth catalan number.
A function to count number of binary trees with n nodes.
find count of BST with n numbers.
return count * n !.
Driver Code.
find count of BST and binary trees with n nodes.
print count of BST and binary trees with n nodes"
Solution in Python:

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What is a code to total number of possible Binary Search Trees and Binary Trees with n keys and the other steps required?
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#Here is code to total number of possible Binary Search Trees and Binary Trees with n keys and the other steps required.
 def factorial ( n ) :
	res = 1
	for i in range ( 1 , n + 1 ) :
		res *= i
	return res
def binomialCoeff ( n , k ) :
	res = 1
	if ( k > n - k ) :
		k = n - k
	for i in range ( k ) :
		res *= ( n - i )
		res //= ( i + 1 )
	return res
def catalan ( n ) :
	c = binomialCoeff ( 2 * n , n )
	return c // ( n + 1 )
def countBST ( n ) :
	count = catalan ( n )
	return count
def countBT ( n ) :
	count = catalan ( n )
	return count * factorial ( n )
if __name__ == ' _ _ main _ _ ' :
	n = 5
	count1 = countBST ( n )
	count2 = countBT ( n )
	print ( " Count ▁ of ▁ BST ▁ with " , n , " nodes ▁ is " , count1 )
	print ( " Count ▁ of ▁ binary ▁ trees ▁ with " , n , " nodes ▁ is " , count2 )