two dimensional array via pointer

Question

I would like to create a dynamic array which store permutation sequence, such that

order[0][]={1,2,3}
order[1][]={2,1,3}
order[2][]={2,3,1}

Answer 1

Source Code:
The code will be something like:

#include 

int **array;
array = malloc(nrows * sizeof(int *));
if(array == NULL)
{
     fprintf(stderr, "out of memory\n");
     /*exit or return*/
}
for(i = 0; i < nrows; i++)
{
    array[i] = malloc(ncolumns * sizeof(int));
    if(array[i] == NULL)
    {
          fprintf(stderr, "out of memory\n");
         /*exit or return*/
    }
}

Concept:

array is a pointer-to-pointer-to-int: at the first level, it points to a block of pointers, one for each row. That first-level pointer is the first one to be allocated; it has nrows elements, with each element big enough to hold a pointer-to-int, or int *. If the allocation is successful then fill in the pointers (all nrows of them) with a pointer (also obtained from malloc) to ncolumns number of ints, the storage for that row of the array.

Pictorial Depiction:

It is simple to grasp if you visualize the situation as:

Taking this into account, the sample code could be rewritten as:

void permute(int num_permute, int num_term, int** order) {
    int x, y;
    int term[5];
    int* ptr = NULL;

    for (y=num_term, x=0; y>0; y--, x++) {
        term[x] = y;
    }
    printf("\n");

    printf("order%12c", ' ');
    for (x=0; x