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How to compute the total number of lottery combinations for choosing 6 out of 37 and 1 power out of 7 in C++

1 Answer

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#include <iostream>
#include <cstdint> // uint64_t

/*
    binomial_coefficient(n, k):
    Computes "n choose k" using the multiplicative formula:

        C(n, k) = product(i = 1..k) of (n - k + i) / i

    Why this method?
        - It avoids huge factorials (e.g., 37! is far too large for 64-bit)
        - It keeps intermediate values small and exact
        - It is efficient, clean, and idiomatic in C++

    Returns:
        The binomial coefficient as std::uint64_t.
*/
std::uint64_t binomial_coefficient(unsigned n, unsigned k) {
    if (k > n) return 0;

    // Use symmetry: C(n, k) == C(n, n-k)
    if (k > n - k)
        k = n - k;

    std::uint64_t result = 1;

    for (unsigned i = 1; i <= k; ++i) {
        result = result * (n - k + i) / i;
    }

    return result;
}

int main() {
    unsigned main_n = 37;
    unsigned main_k = 6;

    unsigned power_n = 7;
    unsigned power_k = 1;

    // Compute combinations
    std::uint64_t main_combos = binomial_coefficient(main_n, main_k);
    std::uint64_t power_combos = binomial_coefficient(power_n, power_k);

    std::uint64_t total = main_combos * power_combos;

    std::cout << "Main combinations (C(37,6)): " << main_combos << "\n";
    std::cout << "Power combinations (C(7,1)): " << power_combos << "\n";
    std::cout << "Total lottery combinations: " << total << "\n";
}



/*
run:

Main combinations (C(37,6)): 2324784
Power combinations (C(7,1)): 7
Total lottery combinations: 16273488

*/

 



answered Jul 27 by avibootz

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