一次从对象列表中查找所有排列

Question

假设我有一个字符串（或者它可能是卡片组），我想通过该字符串的N个项目找到每个排列。我实现了这个算法来找到3个字符串项的所有排列。我有3＆＃34; for＆＃34;在那里循环，它看起来很脏。我想制作一个方法，我将传递参数，告诉我想要多少项目进行排列。听起来像递归......递归让我大吃一惊。只是无法找到一种方法来实现它。

到目前为止，这是我为3个项目的所有排列做的（工作正常，但我想根据所有排列应该有多少项来定制）：

class Program
{
    private static string str = "abcde";
    private static int countPermutations;
    private static int countCombinations;

    static void Main(string[] args)
    {
        char[] list = str.ToCharArray();
        GetAllPermutationOfNAtTime(list);
        Console.WriteLine("{0} {1}", countPermutations, "permutations");
        Console.WriteLine("{0} {1}", countCombinations, "combinations");
        Console.ReadKey();
    }

    //here i finding every combination by 3 items and then 
    //find every permutation of each combination
    private static void GetAllPermutationOfNAtTime(char[] list)
    {
        char[] tempList = new char[3];
        for (int i = 0; i <= list.Length - 3; i++)
        {
            tempList[0] = list[i];
            for (int j = 1; j <= list.Length - 3 + 1; j++)
            {
                if (j <= i)
                    continue;
                tempList[1] = list[j];
                for (int k = 2; k <= list.Length - 3 + 2; k++)
                {
                    if (k <= j)
                        continue;
                    tempList[2] = list[k];
                    countCombinations++;
                    Permutation(tempList, 0, tempList.Length - 1);
                }
            }
        }
    }
    private static void Permutation(char[] list, int start, int end)
    {
        if (start == end)
        {
            Console.WriteLine(list);
            countPermutations++;
        }
        else
        {
            for (int i = start; i <= end; i++)
            {
                Swap(list, start, i);
                Permutation(list, start + 1, end);
                Swap(list, start, i);
            }
        }
    }
    private static void Swap(char[] list, int a, int b)
    {
        if (a == b || list[a] == list[b])
            return;
        char temp;
        temp = list[a];
        list[a] = list[b];
        list[b] = temp;
    }
}