你可以使用IteratorRandom来得到一个更短的解，这是迭代器的一个扩展特性，它提供了choose_multiple和choose_multiple_fill这样方便的函数：

use rand::{seq::IteratorRandom, thread_rng}; // 0.6.1

fn main() {
    let mut rng = thread_rng();
    let v = vec![1, 2, 3, 4, 5];
    let sample = v.iter().choose_multiple(&mut rng, 2);

    println!("{:?}", sample);
}

听起来像是可以使用permutate机箱中的排列：

extern crate permutate; // 0.3.2

use permutate::Permutator;
use std::io::{self, Write};

fn main() {
    let stdout = io::stdout();
    let mut stdout = stdout.lock();
    let list: &[&str] = &["one", "two", "three", "four"];
    let list = [list];
    let mut permutator = Permutator::new(&list[..]);

    if let Some(mut permutation) = permutator.next() {
        for element in &permutation {
            let _ = stdout.write(element.as_bytes());
        }
        let _ = stdout.write(b"\n");
        while permutator.next_with_buffer(&mut permutation) {
            for element in &permutation {
                let _ = stdout.write(element.as_bytes());
            }
            let _ = stdout.write(b"\n");
        }
    }
}