It suffices to show the number of ways to divide the points into 2 non-empty parts such that they can be separated by a line. Now for any such division, consider a separating line. We pretend all the points are pins and the line is a very long stick. Now tilt the stick anti-clockwise as much as possible until it is blocked by 2 pins and cannot be tilted anymore. We call the 2 pins blocking points for every division. We note that for every 2 blocking points, we can find a unique division and for every division, we can find 2 unique blocking points. So the number of divisions is the number of pairs of blocking points which is $\frac{n(n-1)}{2}$.

I guess an "envelop" is a "convex hull". If that's the case, here's a solution: If the two subsets have two non-intersecting convex hull, then it means that the whole set is divided into two parts with the help of a line, which acts as a divider. Because in that case they become not only non-intersecting, but also they don't have a common region. Denote, by $f(n),$ the number of ways in which a segment can divide the whole region into two parts. Then, we assume that there are $n$ points $A_1, A_2, \cdots, A_n$ in the plane, and we introduce a new point $A_{n+1}$ into the set. WLOG assume that $\displaystyle\overline{A_nA_{n+1}}=\min_{1\leq i\leq n}\{\overline{A_{i}A_{n+1}}\}.$ Now, we consider two cases. Case 1. $A_{n+1}$ and $A_n$ are together in one group. In this case, the number of such ways of dividing into two subsets is $f(n).$ Case 2. $A_{n+1}$ and $A_n$ are in different groups. In this case, the number of divisions can be obtained by considering a straight line $\ell_1$ which divides $A_{n+1}$ into a separate set from all the other points, and defining our Pivot as $\ell_1\cap \overleftrightarrow{A_{n+1}A_n}.$ Because no three points are collinear, so on rotating $\ell_1$ about our Pivot, it crosses each of the other $n$ points exactly once. We get the $n$ positions of the line $\ell_1,$ and so, the number of possible ways for this case is $n.$ Using this, we obtain $f(n+1)=f(n)+n;$ Which, in accordance with $f(2)=1,$ gives us \[f(n)=1+2+\cdots+n-1= \binom{n}{2}.\] Therefore, the required number of ways is exactly $\displaystyle\boxed{\binom{n}{2}}.\Box$

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My solution:

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