problem 24

"""
permutation is an ordered arrangement of objects. For example, 3124 is one possible permutation of the digits 1, 2, 3 and 4. If all of the permutations are listed numerically or alphabetically, we call it lexicographic order. The lexicographic permutations of 0, 1 and 2 are:

012   021   102   120   201   210

What is the millionth lexicographic permutation of the digits 0, 1, 2, 3, 4, 5, 6, 7, 8 and 9?"""

# first way  to solve  it 
counter  = 0
for i in range(10):
  for j in range(10):
    if(j != i):
     for k in range(10):
       if (k != j and k != i):
         for a in range(10):
           if (a != j and a != k and a != i):
             for b in  range(10):
               if (b != a and b != k and b != i and b != j):
                 for c in  range(10):
                   if (c != a and c != k and c != i and c != j and c != b):
                     for d in  range(10):
                        if( d != a and d != k and d != i and d != j and d != b and d != c):
                          for e in range(10):
                            if( e != a and e != k and e != i and e != j and e != b and e != c and e != d):
                              for f in  range(10):
                                if ( f != a and f != k and f != i and f != j and f != b and f != c and f != d and f != e):
                                  for w in range(10):
                                    if ( w != a and w != k and w != i and w != j and w != b and w != c and w != d and w != e and w != f):
                                      counter += 1
                                      if counter == 1000000:
                                        print(str(i) + str(j) + str(k) + str(a) + str(b) + str(c) + str(d) + str(e) + str(f) + str(w))

# the second one 
def perm(nums, output , remaining_digits):
  if len(remaining_digits) < 1:
    nums.append(output)
  for digit in remaining_digits:
    perm(nums, output + [digit], remaining_digits - set([digit]))


nums = list()
perm(nums, [], set(range(5)))
print(nums)