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    <p>This code works out the variations, then runs the permutations on each unique set of 3.</p> <p>i.e. for "A", "B", "C", "D" the possibilities are [[A, B, C], [A, B, D], [A, C, D], [B, C, D]]. We then calculate the permutations on each threesome (or n-some) and append the possibilities to a list.</p> <p><strong>PermutationsOfN.processSubsets( List set, int k )</strong> returns: [[A, B, C], [A, B, D], [A, C, D], [B, C, D]]</p> <p>Taking it a bit further <strong>PermutationsOfN.permutations( List list, int size )</strong> returns:<br> [[A, B, C], [A, C, B], [C, A, B], [C, B, A], [B, C, A], [B, A, C], [A, B, D], [A, D, B], [D, A, B], [D, B, A], [B, D, A], [B, A, D], [A, C, D], [A, D, C], [D, A, C], [D, C, A], [C, D, A], [C, A, D], [B, C, D], [B, D, C], [D, B, C], [D, C, B], [C, D, B], [C, B, D]]</p> <pre><code>import java.util.Collection; import java.util.List; import com.google.common.collect.Collections2; import com.google.common.collect.ImmutableList; import com.google.common.collect.Lists; public class PermutationsOfN&lt;T&gt; { public static void main( String[] args ) { List&lt;String&gt; f = Lists.newArrayList( "A", "B", "C", "D" ); PermutationsOfN&lt;String&gt; g = new PermutationsOfN&lt;String&gt;(); System.out.println( String.format( "n=1 subsets %s", g.processSubsets( f, 1 ) ) ); System.out.println( String.format( "n=1 permutations %s", g.permutations( f, 1 ) ) ); System.out.println( String.format( "n=2 subsets %s", g.processSubsets( f, 2 ) ) ); System.out.println( String.format( "n=2 permutations %s", g.permutations( f, 2 ) ) ); System.out.println( String.format( "n=3 subsets %s", g.processSubsets( f, 3 ) ) ); System.out.println( String.format( "n=3 permutations %s", g.permutations( f, 3 ) ) ); System.out.println( String.format( "n=4 subsets %s", g.processSubsets( f, 4 ) ) ); System.out.println( String.format( "n=4 permutations %s", g.permutations( f, 4 ) ) ); System.out.println( String.format( "n=5 subsets %s", g.processSubsets( f, 5 ) ) ); System.out.println( String.format( "n=5 permutations %s", g.permutations( f, 5 ) ) ); } public List&lt;List&lt;T&gt;&gt; processSubsets( List&lt;T&gt; set, int k ) { if ( k &gt; set.size() ) { k = set.size(); } List&lt;List&lt;T&gt;&gt; result = Lists.newArrayList(); List&lt;T&gt; subset = Lists.newArrayListWithCapacity( k ); for ( int i = 0; i &lt; k; i++ ) { subset.add( null ); } return processLargerSubsets( result, set, subset, 0, 0 ); } private List&lt;List&lt;T&gt;&gt; processLargerSubsets( List&lt;List&lt;T&gt;&gt; result, List&lt;T&gt; set, List&lt;T&gt; subset, int subsetSize, int nextIndex ) { if ( subsetSize == subset.size() ) { result.add( ImmutableList.copyOf( subset ) ); } else { for ( int j = nextIndex; j &lt; set.size(); j++ ) { subset.set( subsetSize, set.get( j ) ); processLargerSubsets( result, set, subset, subsetSize + 1, j + 1 ); } } return result; } public Collection&lt;List&lt;T&gt;&gt; permutations( List&lt;T&gt; list, int size ) { Collection&lt;List&lt;T&gt;&gt; all = Lists.newArrayList(); if ( list.size() &lt; size ) { size = list.size(); } if ( list.size() == size ) { all.addAll( Collections2.permutations( list ) ); } else { for ( List&lt;T&gt; p : processSubsets( list, size ) ) { all.addAll( Collections2.permutations( p ) ); } } return all; } } </code></pre> <p>A special mention goes to <a href="https://stackoverflow.com/users/183406/meriton">meriton</a> whose answer <a href="https://stackoverflow.com/questions/4504974/how-to-iteratively-generate-k-elements-subsets-from-a-set-of-size-n-in-java">here</a> helped me work it out.</p>
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    1. COHi mrswadge, thanks for the detailed input ! Though your Java skills are way greater than mine and I'm unable to modify your code so I only get the following list : [[A, B, C], [A, B, D], [A, C, D], [B, C, D]] ;).
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    2. COAh, sorry - I see what you mean! I just re-read your question. Change the method processSubsets( list, size ) to public. Call that instead of the method permutations( List<T> list, int size ). Job done :)
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    3. COWonderful, I think I fully understand your solution now. Thank you!
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