Random element of List<T> from LINQ SQL

Like this:

var rand = new Random();
var user = users[rand.Next(users.Count)];

Use ElementAt:

var rand = new Random();
var user = users.ElementAt( rand.Next( users.Count() ) );

Why not create a generic helper and/or extension?!

namespace My.Core.Extensions
{
    public static class EnumerableHelper<E>
    {
        private static Random r;

        static EnumerableHelper()
        {
            r = new Random();
        }

        public static T Random<T>(IEnumerable<T> input)
        {
            return input.ElementAt(r.Next(input.Count()));
        }

    }

    public static class EnumerableExtensions
    {
        public static T Random<T>(this IEnumerable<T> input)
        {
            return EnumerableHelper<T>.Random(input);
        }
    }
}

Usage would be:

        var list = new List<int>() { 1, 2, 3, 4, 5 };

        var output = list.Random();