Five points, no three of which are collinear, are given. What is the least possible value of the numbers of convex polygons whose some corners are from these five points?

$ 
\textbf{(A)}\ 10
\qquad\textbf{(B)}\ 11
\qquad\textbf{(C)}\ 12
\qquad\textbf{(D)}\ 15
\qquad\textbf{(E)}\ 16
$
1. **Calculate the number of triangles:**
   - We need to choose 3 points out of 5 to form a triangle.
   - The number of ways to choose 3 points from 5 is given by the binomial coefficient:
     \[
     \binom{5}{3} = \frac{5!}{3!(5-3)!} = \frac{5 \times 4 \times 3!}{3! \times 2 \times 1} = 10
     \]
   - Therefore, there are 10 triangles.

2. **Calculate the number of quadrilaterals:**
   - We need to choose 4 points out of 5 to form a quadrilateral.
   - The number of ways to choose 4 points from 5 is given by the binomial coefficient:
     \[
     \binom{5}{4} = \frac{5!}{4!(5-4)!} = \frac{5 \times 4!}{4! \times 1} = 5
     \]
   - Therefore, there are 5 quadrilaterals.

3. **Calculate the number of pentagons:**
   - We need to choose all 5 points to form a pentagon.
   - The number of ways to choose 5 points from 5 is given by the binomial coefficient:
     \[
     \binom{5}{5} = \frac{5!}{5!(5-5)!} = 1
     \]
   - Therefore, there is 1 pentagon.

4. **Sum the number of convex polygons:**
   - The total number of convex polygons is the sum of the number of triangles, quadrilaterals, and pentagons:
     \[
     10 \text{ (triangles)} + 5 \text{ (quadrilaterals)} + 1 \text{ (pentagon)} = 16
     \]

5. **Consider the minimum number of convex polygons:**
   - The problem asks for the least possible value of the number of convex polygons.
   - Since no three points are collinear, every set of 3 points forms a triangle, every set of 4 points forms a quadrilateral, and all 5 points form a pentagon.
   - Therefore, the minimum number of convex polygons is indeed the total number calculated above.

The final answer is $\boxed{16}$