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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 Rust

1 Answer

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/*
    This program computes the total number of lottery combinations for:
        - choosing 6 numbers out of 37
        - choosing 1 power number out of 7

    Total combinations = C(37,6) * C(7,1)

    We implement an idiomatic binomial coefficient function using the
    multiplicative formula:

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

    Why this method?
        - Avoids huge factorials (37! is far too large for u64)
        - Keeps intermediate values small and exact
        - Efficient, clean, and idiomatic Rust
*/

fn binomial_coefficient(n: u64, k: u64) -> u64 {
    if k > n {
        return 0;
    }

    // Use symmetry: C(n, k) == C(n, n-k)
    let mut k = k;
    if k > n - k {
        k = n - k;
    }

    let mut result: u64 = 1;

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

    result
}

fn main() {
    let main_n: u64 = 37;
    let main_k: u64 = 6;

    let power_n: u64 = 7;
    let power_k: u64 = 1;

    let main_combos: u64 = binomial_coefficient(main_n, main_k);
    let power_combos: u64 = binomial_coefficient(power_n, power_k);

    let total: u64 = main_combos * power_combos;

    println!("Main combinations (C(37,6)): {}", main_combos);
    println!("Power combinations (C(7,1)): {}", power_combos);
    println!("Total lottery combinations: {}", total);
}


/*
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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