#Permutation  position
In this kata you will have to permutate through a string of lowercase letters, each permutation will start at ```a``` and you must calculate how many iterations it takes to reach the current permutation.

##examples
```
input: 'a'
result: 1

input: 'c'
result: 3

input: 'z'
result: 26

input: 'foo'
result: 3759

input: 'aba'
result: 27

input: 'abb'
result: 28

```

Source: codewars | Difficulty: EASY
Tags: Fundamentals

Starter code: def permutation_position(perm):
	

Solutions:
from functools import reduce

def permutation_position(perm):
	return reduce(lambda t, c: t * 26 + ord(c) - 97, perm, 0) + 1


trans_table = str.maketrans('abcdefghijklmnopqrstuvwxyz', '0123456789abcdefghijklmnop')

def permutation_position(perm):
	return int(perm.translate(trans_table), 26) + 1


def permutation_position(perm):
	return sum(((ord(c) - 97) * 26 ** i for (i, c) in enumerate(reversed(perm)))) + 1


def permutation_position(x):
	return sum(((ord(c) - 97) * 26 ** p for (p, c) in enumerate(x[::-1]))) + 1


def permutation_position(perm):
	return sum(('abcdefghijklmnopqrstuvwxyz'.index(x) * 26 ** i for (i, x) in enumerate(perm[::-1]))) + 1


def permutation_position(perm):
	a = {'a': 1, 'b': 2, 'c': 3, 'd': 4, 'e': 5, 'f': 6, 'g': 7, 'h': 8, 'i': 9, 'j': 10, 'k': 11, 'l': 12, 'm': 13, 'n': 14, 'o': 15, 'p': 16, 'q': 17, 'r': 18, 's': 19, 't': 20, 'u': 21, 'v': 22, 'w': 23, 'x': 24, 'y': 25, 'z': 26}
	perm = perm.lower()
	b = 0
	y = 1
	z = []
	while b < len(perm) - 1:
		z.append((a[perm[b]] - 1) * 26 ** (len(perm) - y))
		b = b + 1
		y = y + 1
	z.append(a[perm[-1]])
	return sum(z)


permutation_position = lambda p, a=__import__('string').printable: int(p.translate(p.maketrans(a[10:36], a[:26])), 26) + 1


permutation_position = lambda p: sum((26 ** i * (ord(c) - 97) for (i, c) in enumerate(p[::-1]))) + 1


def permutation_position(perm):
	a = list(perm)[::-1]
	res = 0
	k = 0
	for el in a:
		res += (ord(el) - 97) * 26 ** k
		k += 1
	return res + 1