Combination Generator in Linq

For what it's worth, try something like this:

public static IEnumerable<string> GetPermutations(string s)
{
    if (s.Length > 1)
        return from ch in s
               from permutation in GetPermutations(s.Remove(s.IndexOf(ch), 1))
               select string.Format("{0}{1}", ch, permutation);

    else
        return new string[] { s };
}

How to use LINQ to find all combinations of n items from a set of numbers?

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)));
    }
}

getting repeated combinations using LINQ

LINQ is excellent for searching; it's much worse at generating, so it is a wrong tool for the job. So is recursion (although it is a better choice than LINQ). What you need is two nested loops, like this:

var elements = new [] {1, 2};
for (var i = 0 ; i != 5 ; i++) {
    for (int j = 0 ; j != 5 ; j++) {
        if (i == j) continue;
        var data = new int[5];
        data[i] = elements[0];
        data[j] = elements[1];
        ProcessCombination(data);
    }
}

Here is what the code does: at each iteration of the inner loop it places the first and the second element from the elements array at distinct positions i and j. Nested loops go from 0 to 5, making sure all position combinations are covered.

Generate all Combinations from Multiple (n) Lists

Hope this helps.

class NListBuilder

{
    Dictionary<int, List<string>> tags = new Dictionary<int, List<string>>();

    public NListBuilder()
    {
        tags.Add(1, new List<string>() { "A", "B", "C" });
        tags.Add(2, new List<string>() { "+", "-", "*" });
        tags.Add(3, new List<string>() { "1", "2", "3" });
    }

    public List<string> AllCombos
    {
        get
        {
            return GetCombos(tags);
        }
    }

    List<string> GetCombos(IEnumerable<KeyValuePair<int, List<string>>> remainingTags)
    {
        if (remainingTags.Count() == 1)
        {
            return remainingTags.First().Value;
        }
        else
        {
            var current = remainingTags.First();
            List<string> outputs = new List<string>();
            List<string> combos = GetCombos(remainingTags.Where(tag => tag.Key != current.Key));

            foreach (var tagPart in current.Value)
            {
                foreach (var combo in combos)
                {
                    outputs.Add(tagPart + combo);
                }
            }

            return outputs;
        }

    }
}

Select all unique combinations of a single list, with no repeats, using LINQ

Sure - you can do it in a single call to SelectMany with an embedded call to Skip:

var query = slotIds.SelectMany((value, index) => slotIds.Skip(index + 1),
                               (first, second) => new { first, second });

Here's an alternative option, which doesn't use quite such an esoteric overload of SelectMany:

var query = from pair in slotIds.Select((value, index) => new { value, index })
            from second in slotIds.Skip(pair.index + 1)
            select new { first = pair.value, second };

These do basically the same thing, just in slightly different ways.

Here's another option which is much closer to your original:

var query = from index in Enumerable.Range(0, slotIds.Count)
            let first = slotIds[index] // Or use ElementAt
            from second in slotIds.Skip(index + 1)
            select new { first, second };

How to select distinct based on combination of two fields with LINQ?

You can select just the name and age into an anonymous type, and then use Distinct():

var results = context.people
                     .Select(p => new { p.Name, p.Age })
                     .Distinct();

Obviously that won't give you the height, but then there is no single height. If you want all the heights as well, you need to group instead:

var results = context.people
                     .GroupBy(p => new { p.Name, p.Age });

That will give you a sequence of groupings, where each grouping contains all the people with the same name and age.