Yesterday a member of COS asked how to generate all the elements of {*all sets consisting of arbitrary combinations of n elements*} in R and I didn’t know the function *combn()* then, so after a while I constructed a simple loop to fulfill this function (later I found this was wrong because it didn’t enumerate all possible combinations!):

```
> x = 1:5 # as an example, you may change this vector
> n = length(x)
> for (i in 1:n) {
if (i == 1) {
for (j in 1:n) {
print(x[j])
}
}
else {
for (j in 1:(n - i + 1)) {
for (k in (j + i - 1):n) {
print(c(x[j:(j + i - 2)], x[k]))
}
}
}
}
```

The output is:

```
[1] 1
[1] 2
[1] 3
[1] 4
[1] 5
[1] 1 2
[1] 1 3
[1] 1 4
[1] 1 5
[1] 2 3
[1] 2 4
[1] 2 5
[1] 3 4
[1] 3 5
[1] 4 5
[1] 1 2 3
[1] 1 2 4
[1] 1 2 5
[1] 2 3 4
[1] 2 3 5
[1] 3 4 5
[1] 1 2 3 4
[1] 1 2 3 5
[1] 2 3 4 5
[1] 1 2 3 4 5
```

Just now when I was reading the help text of the function *choose()*, I found that there had already been a function *combn()* in the package utils, which is very appropriate for this situation. For example:

```
> combn(x = 5, m = 3)
[,1] [,2] [,3] [,4] [,5] [,6] [,7] [,8] [,9] [,10]
[1,] 1 1 1 1 1 1 2 2 2 3
[2,] 2 2 2 3 3 4 3 3 4 4
[3,] 3 4 5 4 5 5 4 5 5 5
```

So the problem in the beginning will be much easier.