Get nth permutation of a list, while permutations are created in a lexicographical order.

perm.py

#!/bin/env python
""" 
Solves a problem of getting permutation of a given index, 
while you know that permutations are created in a lexicographical order. 
As input you are specifying a list of elements and index of permutation you want to get.
"""

## imports 
from math import factorial
## this one just for checking
from itertools import permutations

def getNthPermutation( pIndex, inList, outList=[] ):
  """ get :pIndex: permutation of :inList:
 
  :warn: permutations are sorted in lexicographical order starting from 0, i.e.:
  0 -> [1, 2, 3], 1 -> [1, 3, 2] and so on

  :param int pIndex: given permutation index
  :param list inList: initial list of elements
  :param list outList: result list
  """
  ## permutation index too big
  if pIndex >= factorial( len(inList) ): return []
  ## no more elements to use
  if not inList: return outList
  ## make sure eList is sorted
  inList.sort()
  ## get factorial
  f = factorial( len(inList)-1 )
  index = pIndex / f
  reminder = pIndex % f
  ## add new element to outList
  outList.append( inList[index] )
  ## ...and remove it from inList
  del inList[index]
  ## bail out or call recursively depending of the reminder 
  if not reminder:
    return outList + inList
  else:
    return getNthPermutation( reminder, inList, outList )

### test execution
if __name__ == "__main__":
  ps = permutations( [1,2,3,4,5,6] ) 
  for i in range(factorial(6)):
    resList = []
    res = getNthPermutation( i, [1,2,3,4,5,6], resList )
    check = list( ps.next() )
    assert( res == check )

What about of the form nPr = n!/(n-r)! , Your current code is n=r ... Thanks in advance