C# String permutation.

[Dasharnb777]

I have 5 strings, such as: "one", "two", "three", "four", and "five". I need to get all permutations of these words. I've explored all internet resources, but all solutions are so bulky and it's hard for me to understand it and integrate it to my program.
So, maybe you know any easy solution how to get permutations.
For example string are 1 2 3, it would: 123, 132, 213, 231, 312, 321.
Thanx.

[Mitja Bonca]

private static void Test()
        {
            List<string> list = new List<string>() { "A", "B", "C" };
            IList<IList<string>> perms = Permutations(list);
            foreach (IList<string> perm in perms)
            {
                foreach (string s in perm)
                {
                    Console.Write(s);
                }
                Console.WriteLine();
            }
        }

        private static IList<IList<T>> Permutations<T>(IList<T> list)
        {
            List<IList<T>> perms = new List<IList<T>>();

            if (list.Count == 0)
                return perms; // Empty list.

            int factorial = 1;
            for (int i = 2; i <= list.Count; i++)
                factorial *= i;

            for (int v = 0; v < factorial; v++)
            {
                List<T> s = new List<T>(list);                
                int k = v;
                for (int j = 2; j <= list.Count; j++)
                {
                    int other = (k % j);
                    T temp = s[j - 1];
                    s[j - 1] = s[other];
                    s[other] = temp;

                    k = k / j;
                }
                perms.Add(s);
            }

            return perms;
        }

The printout from the Test method is:
CAB
CBA
BCA
ACB
BAC
ABC

[Momerath]

Take a look at http://www.daniweb.com/code/snippet349081.html

[Dasharnb777]

Thanx a lot, but as it turned out I need combinations without permutaion. But I'll collect your answer.

[Mitja Bonca]

So mark this thread as answered (cause i beleive it was) and start a new one -if necessary.
Mitja

[Dasharnb777]

So mark this thread as answered (cause i beleive it was) and start a new one -if necessary.
Mitja

Thanx, I want to make it solved but I can't find how to do it.

[Mitja Bonca]

There is a button "Mark as Answered" I think.

[Dasharnb777]

Oh, sorry I can't find it!

[Momerath]

Just FYI Combination Generation is right here