Using recursion we can solve this. I am giving the pesudo code with a small example. let the string be "abc". I am showing the recursive call within the bracket itself. Just to understand. Don't confuse that this is a code inside permute function. permute(abc) { ----level 1----- letter saved at this level is = a; all combination List obtained at this level = permute(bc); after the call returns put the saved letter {a} at each position in each entry and do the same for all entry in the list List obtained here is only bc and cb; that means put a every position of the return bc and cb ----> for bc {abc,back,bcd} and for cb {cab,cab,cba} return all the entries. At the highest level you will get all the permutation of the string. --------------level 2----------- letter saved = b ; all combination List obtained at this level = permute(c); after the call returns put the saved letter at each position in each entry and do the same for all entry in the list List obtained here is only c; that means put b every position of the return c --> bc{putting b before c} and cb{putting b after c} return bc and cb. --------------level 3----------- letter saved = c ; all combination List obtained at this level = 1; //Here it only c is returned Last letter is returned as this is only 1 combination exist -- base condition of recursion. }

C# code. I'm a tester, so my code is a bit rusty, but the algorithm works. public static void Permutations(string orgStr) { List permutations = new List(); _perm(orgStr.ToCharArray(), String.Empty, ref permutations); foreach (string str in permutations) { Console.WriteLine(str); } } private static void _perm(char[] sample, string filled, ref List permutations) { if (sample.Length == 1) { string perm = String.Format("{0}{1}", filled, new string(sample)); permutations.Add(perm); } else { for (int i = 0; i i) { newSample[j-1] = sample[j]; } else { newSample[j] = sample[j]; } } _perm(newSample, newFilled, ref permutations); } } }

# include # include /* Function to swap values at two pointers */ void swap (char *x, char *y) { char temp; temp = *x; *x = *y; *y = temp; } /* Function to print permutations of string This function takes three parameters: 1. String 2. Starting index of the string 3. Ending index of the string. */ void permute(char *a, int i, int n) { int j; if (i == n) printf("%s\n", a); else { for (j = i; j <= n; j++) { swap((a+i), (a+j)); permute(a, i+1, n); swap((a+i), (a+j)); //backtrack } } } /* Driver program to test above functions */ int main() { char a[] = "ABC"; permute(a, 0, 2); getchar(); return 0; }