Question

How to compute the total number of combinations when choosing 7 numbers out of 33 in C

Answer 1

#include <stdio.h>

/*
 * Compute the number of combinations C(n, r):
 *
 *     C(n, r) = n! / (r! * (n - r)!)
 *
 * The calculation is performed incrementally to avoid
 * computing large factorials.
 */
unsigned long long combinations(unsigned int n, unsigned int r)
{
    /* Use symmetry: C(n, r) == C(n, n - r) */
    if (r > n - r) {
        r = n - r;
    }

    unsigned long long result = 1;

    /*
     * Compute:
     *
     * (n     )   (n - 1)   ...   (n - r + 1)
     * -------- * ------- * ... * -----------
     *    1          2                 r
     */
    for (unsigned int i = 0; i < r; ++i) {
        result *= (n - i);
        result /= (i + 1);
    }

    return result;
}

int main(void)
{
    const unsigned int total_numbers = 33;
    const unsigned int chosen_numbers = 7;

    unsigned long long total_combinations =
        combinations(total_numbers, chosen_numbers);

    printf("C(%u, %u) = %llu\n",
           total_numbers,
           chosen_numbers,
           total_combinations);

    return 0;
}


/*
run:

C(33, 7) = 4272048

*/

 

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