Calculating all string permutations of a given string can be solved in O(n!) by trying all possibilities.
Now, looking at the Travel Salesman Problem, we can solve it by trying all permutations of cities. Lets say we have cities A, B and C. Lets say we start at city A. By calculating all permutations of BC string we get ABC ACB, then we just sum (in polynomial time the distance between AB, CB and CA for the first case...)
So isnt this a reduction of the all strings permutation to the travel salesman problem and isnt it a NP Complete problem?