hi , this is the output of the program ;Code:#include <iostream>
#include <algorithm>
using namespace std;
int main () {
int n=4;
int myints[] = {1,2,3,4};
do {
for(int i=0 ; i<n ;i++)
cout << myints[i];
cout<<endl;
} while ( next_permutation (myints,myints+2) );
return 0;
}
1 2 3 4
2 1 3 4
now the question is how can i make the output like this ;
1 2 3 4
1 2 4 3
2 1 3 4
2 1 4 3
if you realize first 2 number had made a permutation each other and the other last 2 number had a permutation each other ,
the example is about 4 number making and per 2 of them making permutation ( 2!*2!= 4 possible results )
from here if you give me the idea i will try to make this code for ' n '
like we say there are 9 number and per 3 of them makes permutation,(3! *3!*3! = 108 possible results)