Snake permutations.py

""" Given the first 10 numbers, we want to know how many snake permutations exist, where a snake permutations is:
    x1 < x2 > x3 < x4 > x5 < x6 > x7 < x8 > x9 < x10 
    The result is very counterintuitive....more than 50000 snake permutation exist.
    It takes a lot of time to have the output, so take only the idea of the code..."""
import random

union = []
for i in range (25000000):
    A = [s for s in range (1,11)]
    new = []
    for i in range(10):
        k = random.choice(A)
        new.append(k)
        A.remove(k)
    count = 0
    for i in range(0,8,2):
        if new[i] < new[i+1] and new[i+1] > new[i+2] and new[8] < new[9]:  # the algorithm to search the snake permutaton
            count += 1
    if count == 4 and new not in union:  
        union.append(new)

print(union)
print()
print("For n = 10 there are: {} snakes permutations".format(len(union)))