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Count of Binary Strings of length N such that frequency of 1 ' s ▁ exceeds ▁ frequency ▁ of ▁ 0' s | Function to calculate and return the value of Binomial Coefficient C ( n , k ).
Since C ( n , k ) = C ( n , n - k ).
Calculate the value of [ n * ( n - 1 ) * -- - * ( n - k + 1 ) ] / [ k * ( k - 1 ) * -- - * 1 ].
Function to return the count of binary strings of length N such that frequency of 1 ' s ▁ exceed ▁ that ▁ of ▁ 0' s.
Count of N - length binary strings.
Count of N - length binary strings having equal count of 0 ' s ▁ and ▁ 1' s.
For even length strings.
Driver Code"
Solution in Python:
assistant
#Here is code to count of Binary Strings of length N such that frequency of 1 ' s ▁ exceeds ▁ frequency ▁ of ▁ 0' s and the other steps required.
 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 countOfString ( N ) :
	Stotal = pow ( 2 , N )
	Sequal = 0
	if ( N % 2 == 0 ) :
		Sequal = binomialCoeff ( N , N // 2 )
	S1 = ( Stotal - Sequal ) // 2
	return S1
N = 3
print ( countOfString ( N ) )