# løsning 1
def badminton(per_vil, palle_vil, espen_vil):
    if (per_vil and palle_vil and not espen_vil) or (per_vil and espen_vil and not palle_vil) or (espen_vil and palle_vil and not per_vil):
        return True
    return False



# løsning 2
def badminton(per_vil, palle_vil, espen_vil):
    return (per_vil + palle_vil + espen_vil) == 2