We have m objects on a circle. We want to take n objects of it ($n\le m$ ). Remark: taking the 1st and 3nd object is identical as taking the 4th and 6th object. The answer is $\frac{\binom{m}{n}}{m}$.
Why do we divide by m ? (ok, we can pick up m 'lines' like this on the circle, but I don't get it)
Note this approach assumes the original $m$ objects are equally spaced around the circle. But that's usually assumed in these circular type arrangement problems, certainly when dividing by $m$ comes into play.
Since any choice can be rotated into $m$ different positions I see why one might think to divide by $m.$ But I don't think that's right since some positions rotate to more than one, like even $m$ and pick every other one.
Added: I think it works if $m$ happens to be prime, since then any position (subset of size $n$) is uniquely determined mod $m.$
Oops doesn't work for primes either. If $m=n=7$ the formula would give $1/7.$ But it works if $m$ prime and $1<n<m.$