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    <p>Following recursive logic allows you to calculate all combinations without repetitions and without the need to calculate all of them first. It works pretty nice, as long as choose(nx-1,ning-1) returns an integer. If it doesn't, calculating the possibilities is a bit ridiculous.</p> <p>It's a recursive process, so it might take long and it will cause memory trouble when your vectors exceed a certain limit. But then again, dividing a set of 14 elements in 7 groups gives already 135135 unique possibilities. Things get out of hand pretty quick in these kind of things.</p> <p>The logic in pseudo-something (wouldn't call it pseudocode)</p> <pre><code>nb = number of groups ning = number of elements in every group if(nb == 2) 1. take first element, and add it to every possible combination of ning-1 elements of x[-1] 2. make the difference for each group defined in step 1 and x to get the related second group 3. combine the groups from step 2 with the related groups from step 1 if(nb &gt; 2) 1. take first element, and add it to every possible combination of ning-1 elements of x[-1] 2. to define the other groups belonging to the first groups obtained like this, apply the algorithm on the other elements of x, but for nb-1 groups 3. combine all possible other groups from step 2 with the related first groups from step 1 </code></pre> <p>Translating this to R gives us :</p> <pre><code>perm.groups &lt;- function(x,n){ nx &lt;- length(x) ning &lt;- nx/n group1 &lt;- rbind( matrix(rep(x[1],choose(nx-1,ning-1)),nrow=1), combn(x[-1],ning-1) ) ng &lt;- ncol(group1) if(n &gt; 2){ out &lt;- vector('list',ng) for(i in seq_len(ng)){ other &lt;- perm.groups(setdiff(x,group1[,i]),n=n-1) out[[i]] &lt;- lapply(seq_along(other), function(j) cbind(group1[,i],other[[j]]) ) } out &lt;- unlist(out,recursive=FALSE) } else { other &lt;- lapply(seq_len(ng),function(i) matrix(setdiff(x,group1[,i]),ncol=1) ) out &lt;- lapply(seq_len(ng), function(i) cbind(group1[,i],other[[i]]) ) } out } </code></pre> <p>To show it works :</p> <pre><code>&gt; perm.groups(1:6,3) [[1]] [,1] [,2] [,3] [1,] 1 3 5 [2,] 2 4 6 [[2]] [,1] [,2] [,3] [1,] 1 3 4 [2,] 2 5 6 [[3]] [,1] [,2] [,3] [1,] 1 3 4 [2,] 2 6 5 [[4]] [,1] [,2] [,3] [1,] 1 2 5 [2,] 3 4 6 [[5]] [,1] [,2] [,3] [1,] 1 2 4 [2,] 3 5 6 [[6]] [,1] [,2] [,3] [1,] 1 2 4 [2,] 3 6 5 [[7]] [,1] [,2] [,3] [1,] 1 2 5 [2,] 4 3 6 [[8]] [,1] [,2] [,3] [1,] 1 2 3 [2,] 4 5 6 [[9]] [,1] [,2] [,3] [1,] 1 2 3 [2,] 4 6 5 [[10]] [,1] [,2] [,3] [1,] 1 2 4 [2,] 5 3 6 [[11]] [,1] [,2] [,3] [1,] 1 2 3 [2,] 5 4 6 [[12]] [,1] [,2] [,3] [1,] 1 2 3 [2,] 5 6 4 [[13]] [,1] [,2] [,3] [1,] 1 2 4 [2,] 6 3 5 [[14]] [,1] [,2] [,3] [1,] 1 2 3 [2,] 6 4 5 [[15]] [,1] [,2] [,3] [1,] 1 2 3 [2,] 6 5 4 </code></pre>
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    1. POFind all possible ways to split a list of elements into a a given number of group of the same size
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    1. COAwesome ! It seems to be working perfectly ! I'd spent a whole day on that without succeeding ! Thank you and Thanks to all of you dear stackoverflow readers !
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    2. COYou're welcome. You know that you can accept the correct answer by clicking the V-sign on the left of the post?
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    3. COYou've just taught me two things in less than 10 minutes ;)
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