If I have a list of integers:
List myValues = new List(new int[] { 1, 2, 3, 4, 5, 6 } );
How would I get 3 random integer
Combining the other answer with this answer can lead you to the following:
var rand = new Random(); var numbers = Enumerable.Range(1, 6).OrderBy(i => rand.Next()).ToList();
In this case 1 is the starting value (inclusive) and 6 is the number of integers to generate.
1
6