Select N random elements from a List<T> in C#

Using linq:

YourList.OrderBy(x => rnd.Next()).Take(5)

Iterate through and for each element make the probability of selection = (number needed)/(number left)

So if you had 40 items, the first would have a 5/40 chance of being selected. If it is, the next has a 4/39 chance, otherwise it has a 5/39 chance. By the time you get to the end you will have your 5 items, and often you'll have all of them before that.

This technique is called selection sampling, a special case of Reservoir Sampling. It's similar in performance to shuffling the input, but of course allows the sample to be generated without modifying the original data.

public static List<T> GetRandomElements<T>(this IEnumerable<T> list, int elementsCount)
{
    return list.OrderBy(arg => Guid.NewGuid()).Take(elementsCount).ToList();
}