Usage:
var results = new[] { 1, 2, 3, 4, 5, 6, 7, 8, 9 }.DifferentCombinations(3);
Code:
public static class Ex
{
public static IEnumerable<IEnumerable<T>> DifferentCombinations<T>(this IEnumerable<T> elements, int k)
{
return k == 0 ? new[] { new T[0] } :
elements.SelectMany((e, i) =>
elements.Skip(i + 1).DifferentCombinations(k - 1).Select(c => (new[] {e}).Concat(c)));
}
}