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    copied!<p>Here's How to do it using depth first Recursion. Takes about 3 seconds to run on my machine. Also this will work for an arbitrary sized pairing by changing the PAIRCOUNT to say 5 if you had 5 columns instead of 4 and just .add the additional Pairs as appropriate. </p> <pre><code> void Main() { var OriginValues = new List&lt;KeyValuePair&lt;char, int&gt;&gt;(); OriginValues.Add(new KeyValuePair&lt;char, int&gt;('Q', 100)); OriginValues.Add(new KeyValuePair&lt;char, int&gt;('R', 150)); OriginValues.Add(new KeyValuePair&lt;char, int&gt;('W', 250)); OriginValues.Add(new KeyValuePair&lt;char, int&gt;('T', 30)); OriginValues.Add(new KeyValuePair&lt;char, int&gt;('Q', 90)); OriginValues.Add(new KeyValuePair&lt;char, int&gt;('R', 130)); OriginValues.Add(new KeyValuePair&lt;char, int&gt;('W', 225)); OriginValues.Add(new KeyValuePair&lt;char, int&gt;('T', 28)); OriginValues.Add(new KeyValuePair&lt;char, int&gt;('Q', 80)); OriginValues.Add(new KeyValuePair&lt;char, int&gt;('R', 110)); OriginValues.Add(new KeyValuePair&lt;char, int&gt;('W', 210)); OriginValues.Add(new KeyValuePair&lt;char, int&gt;('T', 25)); ///... and the other 7 var AllPermutation = new List&lt;List&lt;KeyValuePair&lt;char, int&gt;&gt;&gt;(); Recurse(OriginValues, ref AllPermutation); //all results will be in AllPermutation now } const int PAIRCOUNT = 4; void Recurse(List&lt;KeyValuePair&lt;char, int&gt;&gt; OriginValues, ref List&lt;List&lt;KeyValuePair&lt;char, int&gt;&gt;&gt; result, List&lt;KeyValuePair&lt;char, int&gt;&gt; itemset = null) { itemset = itemset ?? new List&lt;KeyValuePair&lt;char, int&gt;&gt;(); var temp = new List&lt;KeyValuePair&lt;char, int&gt;&gt;(itemset); if (itemset.Count == OriginValues.Count / PAIRCOUNT) { result.Add(temp); return; } for (int x = 0; x &lt; PAIRCOUNT; x++) { temp.Add(OriginValues[itemset.Count * PAIRCOUNT + x]); Recurse(OriginValues, ref result, temp); temp = new List&lt;KeyValuePair&lt;char, int&gt;&gt;(itemset); } } </code></pre>
 

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