human
N digit numbers divisible by 5 formed from the M digits | Function to find the count of all possible N digit numbers which are divisible by 5 formed from M digits.
If it is not possible to form n digit number from the given m digits without repetition.
If both zero and five exists.
Remaining N - 1 iterations.
Remaining N - 1 iterations.
Driver code"
How can the above be solved in Python?

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Can you provide a code to n digit numbers divisible by 5 formed from the M digits and the other steps required?
assistant
#Here is code to n digit numbers divisible by 5 formed from the M digits and the other steps required.
 def numbers ( n , arr , m ) :
	isZero = 0
	isFive = 0
	result = 0
	if ( m < n ) :
		return - 1
	for i in range ( m ) :
		if ( arr [ i ] == 0 ) :
			isZero = 1
		if ( arr [ i ] == 5 ) :
			isFive = 1
	if ( isZero and isFive ) :
		result = 2
		for i in range ( n - 1 ) :
			m -= 1
			result = result * ( m )
	elif ( isZero or isFive ) :
		result = 1
		for i in range ( n - 1 ) :
			m -= 1
			result = result * ( m )
	else :
		result = - 1
	return result
if __name__ == " _ _ main _ _ " :
	n = 3
	m = 6
	arr = [ 2 , 3 , 5 , 6 , 7 , 9 ]
	print ( numbers ( n , arr , m ) )