Even/Odd permutation

[braankoo]

def is_even(p):
  i = 0
  count = 0
  while(i < len(p)):
    k = i + 1
    while(k < len(p)):
      if(p[i] > p[k]):
        count+=1
      k+=1
    i+=1
  if(count == 0):
    return True
  if(count % 2 == 0):
    return True
  else:
    return False

Here is some solution. Could it be done better? Yes.
:)

This is my first code in python ever, so I was kind of 'scared'.