A permutation on permutations

[Arthur Siegel]

What I've come up with is:

L is my list of arbitrary length.
M= [i for i in range (len(L))]

#thanks to Rainer Deyke's post
def perms(L):
   if list == [] :
      return [[]]
   return [[list[i] + p for i in range(len(list))\
     for p in perms(list[:i] + list[i+1:])]

def removedups(t):
   for p in t:
      m=copy.copy(p)
      m.reverse()
      if m in t:
         t.remove(m)

def drawcurve(t):
   #t is  M or a slice of M
   perm=perms(t)
   removedups(perm)
   for i in range(len(perm)):
      cp=[L[j] for j in perm[i]]
      BezierCurve(cp) 

It seems to work.

But I wonder if there is a significantly more
efficient solution

Art