[CheckIO/Electronic Station] Acceptable Password V

1. Problem: https://py.checkio.org/en/mission/acceptable-password-v/
2. My solution

def is_acceptable_password(password: str) -> bool:
    # your code here
    
    if len(password) < 6: return False
    elif 'password' in password.lower(): return False
    elif  len(password) > 9: return True
    else:
        list_boolen = [ x.isnumeric() for x in list(password)]
        if True in list_boolen : 
            if False in list_boolen: return True
        
        return False