You can make in manually, using the fact that n-bit binary number naturally corresponds to a subset of n-element set.
private IEnumerable<int> constructSetFromBits(int i)
{
for (int n = 0; i != 0; i /= 2, n++)
{
if ((i & 1) != 0)
yield return n;
}
}
List<string> allValues = new List<string>()
{ "A1", "A2", "A3", "B1", "B2", "C1" };
private IEnumerable<List<string>> produceEnumeration()
{
for (int i = 0; i < (1 << allValues.Count); i++)
{
yield return
constructSetFromBits(i).Select(n => allValues[n]).ToList();
}
}
public List<List<string>> produceList()
{
return produceEnumeration().ToList();
}