Question

How to generate all combinations of 3 digits without repetition in any order (012 and 021 are one combination) with C

Answer 1

#include <stdio.h>
 
int main(void) {

    for (int i = 0; i <= 9; i++) {
        for (int j = i + 1; j <= 9; j++) {
            for (int k = j + 1; k <= 9; k++) {
                printf("%d %d %d | ", i, j, k);
            }
            printf("\n");
        }
    }  
     
    return 0;
}
 
 
 
   
/*
run:
   
0 1 2 | 0 1 3 | 0 1 4 | 0 1 5 | 0 1 6 | 0 1 7 | 0 1 8 | 0 1 9 | 
0 2 3 | 0 2 4 | 0 2 5 | 0 2 6 | 0 2 7 | 0 2 8 | 0 2 9 | 
0 3 4 | 0 3 5 | 0 3 6 | 0 3 7 | 0 3 8 | 0 3 9 | 
0 4 5 | 0 4 6 | 0 4 7 | 0 4 8 | 0 4 9 | 
0 5 6 | 0 5 7 | 0 5 8 | 0 5 9 | 
0 6 7 | 0 6 8 | 0 6 9 | 
0 7 8 | 0 7 9 | 
0 8 9 | 

1 2 3 | 1 2 4 | 1 2 5 | 1 2 6 | 1 2 7 | 1 2 8 | 1 2 9 | 
1 3 4 | 1 3 5 | 1 3 6 | 1 3 7 | 1 3 8 | 1 3 9 | 
1 4 5 | 1 4 6 | 1 4 7 | 1 4 8 | 1 4 9 | 
1 5 6 | 1 5 7 | 1 5 8 | 1 5 9 | 
1 6 7 | 1 6 8 | 1 6 9 | 
1 7 8 | 1 7 9 | 
1 8 9 | 

2 3 4 | 2 3 5 | 2 3 6 | 2 3 7 | 2 3 8 | 2 3 9 | 
2 4 5 | 2 4 6 | 2 4 7 | 2 4 8 | 2 4 9 | 
2 5 6 | 2 5 7 | 2 5 8 | 2 5 9 | 
2 6 7 | 2 6 8 | 2 6 9 | 
2 7 8 | 2 7 9 | 
2 8 9 | 

3 4 5 | 3 4 6 | 3 4 7 | 3 4 8 | 3 4 9 | 
3 5 6 | 3 5 7 | 3 5 8 | 3 5 9 | 
3 6 7 | 3 6 8 | 3 6 9 | 
3 7 8 | 3 7 9 | 
3 8 9 | 

4 5 6 | 4 5 7 | 4 5 8 | 4 5 9 | 
4 6 7 | 4 6 8 | 4 6 9 | 
4 7 8 | 4 7 9 | 
4 8 9 | 

5 6 7 | 5 6 8 | 5 6 9 | 
5 7 8 | 5 7 9 | 
5 8 9 | 

6 7 8 | 6 7 9 | 
6 8 9 | 

7 8 9 | 
 
*/

 

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