One solution to the issue of finding the amount of combinations for k elements in a list of length n. (See http://en.wikipedia.org/wiki/Combination#Enumerating_k-combinations).

k_elements_from_n.c

// k_elements_from_n
// c99
#include <stdio.h>
#include <stdlib.h>

unsigned long factorial(long n) {
	return n==1 ? 1 : n*factorial(n-1);
}

void print_arr(int arr[], int length) {
	for (int i=0, l=length; i<l; i++)
		printf("%s%d%s", i==0?"[":"", arr[i], i+1<l?", ":"]\n");
}

// Combo of k elements from n-list; results are printed out
void comb(int k, int n) {
	if (k > n) return;
	int group[k];

	printf("Task: find %d items from a list of %d elements.\n", k, n);
	printf("There are %d!/(%d!%d!) = %lu possible combinations.\n",
	       n, k, n-k, factorial(n)/(factorial(k)*factorial(n-k)));

	// initialize group
	for (int i=0; i<k; i++)
		group[i] = i+1;

	int iter = 0;
	printf("%d. ", ++iter);
	print_arr(group, k);

	for (int p=k-1; p>=0;) { // e[p] at limit
		if (group[p] == n-(k-p-1)) p--;
		else if (p+1<k && group[p+1] == n-(k-p-2)) {
			group[p]++; // e[p] not at limit, but e[p+1] is
			group[p+1] = group[p];
			p++;
		} else {
			group[p]++; // neither; continue to increment e[p]
			printf("%d. ", ++iter);
			print_arr(group, k);
		}
	}
}

int main(int argc, char* argv[]) {
	if (argc > 2)
		for (int a=1; a+1<argc; a+=2)
			comb(atoi(argv[a]), atoi(argv[a+1]));
	else comb(4, 12); // example
	return 0;
}