Question: I want to find out if one string is equal to at least one of the permutations of the other string.
For example, the output of the following two comparisons must be `true`:

```
print('united states of America' == 'America united states of')
print('united states of America' == 'states of united America')

```

I used the `itertools.permutations` method to find permutations, but
I want it for words not the sentence nor the letters in the string.

So not `'abcd' == 'bcad'` equals `True`, but `'a b c d' == 'b c a d'`
should be `True`.

Answer: 
Just check the words frequencies

```
from collections import Counter

c = Counter('united states of America')
c1 = Counter('America united states of')
c2 = Counter('states of united America'

print(c == c1 == c2)
# True

```
If `itertools.permutations` is not a must, I would use sets:

```
>>> set('united states of America'.split()) == set('America united states of'.split())
True
>>> set('united states of America'.split()) == set('America united states off'.split())
False

```