Solving "number of cases" math problem in python

number_of_cases.py

from itertools import permutations

li_Consanant = [ 'B', 'C', 'D', 'F', 'G', 'H']
li_Vowel = ['A', 'E']

# check every consanant is neighbor in a statement

def isValid(statement): # statement list of code
    def isConsanant(code):
        global li_Consanant
        global li_Vowel
        return True if code in li_Consanant else False

    if len(statement) <= 1:
        return False

    # cons_cnt records how many times changed
    # from vowel to consanant in a statement

    last_char = isConsanant(statement[0])
    cons_cnt = 1 if last_char else 0

    for code in statement[1:]:
        cons_state = isConsanant(code)
        if cons_state != last_char and cons_state == True:
            cons_cnt += 1

        last_char = cons_state

    if cons_cnt <= 1:
        return True
    else:
        return False

# execute
validnum = 0
for i in permutations('ABCDEFGH'):
    if isValid(i):
        validnum += 1

print(validnum)

# 3! x 6! = 4320 is the answer of this problem.