'n'字符串的所有可能组合，重复：C= n！/(n-k)！

我想有这样的东西:

combination no 1: sentence1 sentence2 sentence3 sentence4
combination no 2: sentence1 sentence2 sentence4 sentence3
combination no 3: sentence1 sentence3 sentence2 sentence4
combination no 4: sentence1 sentence3 sentence4 sentence2
combination no 5: sentence1 sentence4 sentence3 sentence2
combination no 6: sentence1 sentence4 sentence2 sentence3

等等…

现在，使用下面的代码，我如何处理公式中的"k"变量?少了什么吗?同样，这是关于重复的组合所以我认为公式是C= n!/(n-k)!.

// next_permutation example
#include <iostream>     // std::cout
#include <algorithm>    // std::next_permutation, std::sort
#include <string>       // std::string
#include <vector>       // std::vector
int main () {
  std::string sentence1 = " A Sentence number one ";
  std::string sentence2 = " B Sentence number two ";
  std::string sentence3 = " C Sentence number three ";
  std::string sentence4 = " D Sentence number four ";
  // Store all the elements in a container ( here a std::vector)
  std::vector<std::string> myVectorOfStrings;      
  // In the vector we add all the sentences.
  // Note : It is possible to do myVectorOfStrings.push_back("Some sentence");
  myVectorOfStrings.push_back(sentence1);
  myVectorOfStrings.push_back(sentence2);
  myVectorOfStrings.push_back(sentence3);
  myVectorOfStrings.push_back(sentence4);
  // The elements must be sorted to output all the combinations
  std::sort (myVectorOfStrings.begin(),myVectorOfStrings.end());

  std::cout << "The 4! possible permutations with 4 elements:n";
  do {
    //This printing can be improved to handle any number of sentences, not only four.
    std::cout << myVectorOfStrings[0] << ' ' << myVectorOfStrings[1] << ' ' << myVectorOfStrings[2] << ' ' << myVectorOfStrings[3] << 'n';
  } while ( std::next_permutation(myVectorOfStrings.begin(),myVectorOfStrings.end()) );
  std::cout << "After loop: "  << myVectorOfStrings[0] << ' ' << myVectorOfStrings[1] << ' ' << myVectorOfStrings[2] << ' ' << myVectorOfStrings[3] << 'n';
  return 0;
}