Let $P$ be a set of $n\ge 3$ points in the plane, no three of which are on a line. How many possibilities are there to choose a set $T$ of $\binom{n-1}{2}$ triangles, whose vertices are all in $P$, such that each triangle in $T$ has a side that is not a side of any other triangle in $T$?
1. **Base Case:**
   - For \( n = 3 \), the set \( P \) consists of 3 points, and there is exactly one triangle that can be formed. Hence, the answer is clearly 1.

2. **General Case:**
   - Suppose \( n \ge 4 \). We need to choose a set \( T \) of \( \binom{n-1}{2} \) triangles such that each triangle in \( T \) has a side that is not a side of any other triangle in \( T \).

3. **Edge Coloring:**
   - Consider all \( \binom{n}{2} \) edges between two points in \( P \). We color these edges blue or red:
     - Blue edges appear in exactly one triangle.
     - Red edges do not appear in any triangle or appear in multiple triangles.

4. **Counting Edges:**
   - Each triangle has 3 edges, and we need \( 3 \binom{n-1}{2} \) edges in total (counting with multiplicity).
   - Let \( b \) be the number of blue edges and \( r \) be the number of red edges. We have:
     \[
     b + r = \binom{n}{2}
     \]
     \[
     b + r(n-2) = 3 \binom{n-1}{2}
     \]

5. **Solving the Equations:**
   - From the first equation:
     \[
     r = \binom{n}{2} - b
     \]
   - Substitute \( r \) into the second equation:
     \[
     b + (\binom{n}{2} - b)(n-2) = 3 \binom{n-1}{2}
     \]
     \[
     b + \binom{n}{2}(n-2) - b(n-2) = 3 \binom{n-1}{2}
     \]
     \[
     \binom{n}{2}(n-2) = 3 \binom{n-1}{2}
     \]
   - Simplify using the binomial coefficient identities:
     \[
     \frac{n(n-1)}{2}(n-2) = 3 \frac{(n-1)(n-2)}{2}
     \]
     \[
     \frac{n(n-1)(n-2)}{2} = \frac{3(n-1)(n-2)}{2}
     \]
     \[
     n = 3
     \]
   - This confirms that the number of blue edges \( b \) is \( \binom{n-1}{2} \) and the number of red edges \( r \) is \( n-1 \).

6. **Distribution of Edges:**
   - Any point with one blue edge must have at most one red edge. Therefore, any point has either \( n-1 \) red edges or at most 1 red edge.
   - It is not possible for all points to have at most one red edge since their number would be bounded by \( \frac{n}{2} < n-1 \).
   - Thus, one point must have all \( n-1 \) edges red, and these are all the red edges.

7. **Conclusion:**
   - Each such distribution satisfies the conditions, and there are \( n \) choices for \( T \) since it is uniquely determined by choosing the one point in \( P \) with all edges red.

The final answer is \( \boxed{ n } \).