You probably don't want permutations, since order doesn't matter, correct? I'm guessing you want all possible combinations of 5 from your vector of 10 (without replacement). Check out the functions 'permutations' and 'combinations' from package 'gregmisc'. If you want all possible comparisons of 5 samples versus 5 samples you can determine the remaining samples for each random combination of 5 and then remove duplicates with a combination of 'sort' and 'duplicated'. I think the code you want is below. That produces 252 possible combinations of 5 samples from your list of 10 and the 126 possible comparisons of 5 samples versus 5 samples.

```
library(gregmisc)
labels=c("case_1","case_2","case_3","case_4","case_5","control_1","control_2","control_3","control_4","control_5")
#Get all possible combinations of 5 samples (without replacement)
combos=combinations(n=10, r=5, v=labels)
#Get all possible comparisons of 5 samples vs 5 samples
getpair=function(x){return(labels[which(!labels %in% x)])}
combo_pairs=t(apply(combos,1,getpair))
comparisons=cbind(apply(combos,1,paste, collapse=" "),apply(combo_pairs,1,paste, collapse=" "))
comparisons.sort=t(apply(comparisons, 1, sort))
comparisons.uniq=comparisons[!duplicated(comparisons.sort),]
```

A small aside, this question might have been better posed to the statistics stackexchange where it was answered or stackoverflow [R tagged] where it has been addressed many times. It is not directly a bioinformatics problem, although I grant that it comes up in bioinformatics a lot, and my preferred solution is buried pretty deep in those sources.

Either you want combination or permutation, that does not produces 126 possibilities. Can you be more specific, how did you calculate 126 possibilities?

Ahh. I think I get it now. They want all possible 5 sample vs 5 sample comparisons. For each of the 252 possible combinations of 10 choose 5 that I calculated below you would compare those 5 samples to the remaining 5 samples (which are another combination). This pairing of 5 sample combinations will create duplicates (ABCDE vs FGHIJ is the same as FGHIJ vs ABCDE). So, the number of 126 possible pairs of 5 combinations is correct I think. As long as no replacement allowed anywhere.

Yep. 'combn' works too (I suggested combinations from 'gregmisc' package). See my answer below if you also need help removing the duplicate cases. Set "combos=t(combn(labels,5))" and then the rest of the code will work as indicated.

Yep Obi, you got that right, see "Chris Millers" message for how to produce it. Startlingly straightforward once you know how!