Question

Following program is missing one permutation entry.

#include <iostream>
#include <vector>
#include <algorithm>

int main ( int argc, char **argv) {
    std::vector<int> temp;
    temp.push_back(10);
    temp.push_back(2);
    temp.push_back(4);
    temp.push_back(4);

    do {
        std::copy(temp.begin(),temp.end(),std::ostream_iterator<int>(std::cout," "));
        std::cout << std::endl;
    }while ( std::next_permutation (temp.begin(), temp.end()));
}

Following is the output of the program

10 2 4 4
10 4 2 4
10 4 4 2

why it is missing one entry which is

2 4 4 10

Answer 1

This is because that permutation is the first ordering for the list of number that you have.
You would need to sort the original array, then this permutation will be listed as the very first one.

std::vector<int> temp;
temp.push_back(10);
temp.push_back(2);
temp.push_back(4);
temp.push_back(4);
std::sort(temp.begin(),temp.end() );

Alternatively, you could just push the elements in sorted order, but for practical purposes, you should always sort if you want to generate all possible permutations.