Counting polyhedra

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Given $n>6$ points in space are placed in such a way that no three are collinear and no four lie on the same plane. Show number of convex polyhedra with $5$ faces and vertices among the given points is at least $\dbinom{n-4}{3}$. I found a (maybe relevant) thread here but this does not work or I can't use it so please help me out. Thanks in advance for all help.

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Unless I'm missing something, there are no 5-faced polyhedra among the points.

Since no four points are in the same plane, all of the faces must be triangles. Every edge of a convex polyhedron is in 2 faces, so if we have $e$ edges, then $2 e = 3 \cdot 5$, which is absurd (we cannot have 7.5 edges.)