How many types of convex pentahedra exist?
If a polyhedron had a face that is a pentagon or has more than five sides, then the solid would have at least 6 faces. Therefore, the faces can only be triangles and quadrilaterals. Let $e$ be the number of edges, and let the number of triangular faces be $k$. Then the number of quadrilateral faces is $5-k$. Counting the edges of the solid face by face, we get twice the number of edges, that is,

$$
2 e=3 k+4(5-k), \quad \text { from which } \quad k=20-2 e
$$

Therefore, $k$ is even, and since $k \leq 5$, only $k=0,2,4$ are possible.

The case $k=0$ does not provide a solution. Indeed, there is no convex polyhedron with 5 quadrilateral faces. If 4 quadrilateral faces meet at a vertex, the fifth would need to connect to 8 edges, which is impossible. If, however, only 3 quadrilateral faces meet at each vertex, then counting the vertices face by face, the number of vertices would be $\frac{5 \cdot 4}{3}$, which is also impossible, as the result is not an integer.

In the case $k=2$, we are talking about a solid bounded by two triangles and three quadrilaterals. Here, three quadrilateral faces cannot share a vertex, because the two triangles connected by an edge cannot fit their remaining four edges into 6 edges of quadrilaterals. Thus, the two triangles lie "opposite" each other. Such a solid can be, for example, a triangular prism, or generally, a truncated 3-sided pyramid.

In the case $k=4$, a quadrilateral face is connected to a triangular face on each of its sides. This polyhedron is a quadrilateral-based pyramid.

Therefore, there are two types of convex polyhedra with 5 faces.

Remark. Some solvers used Euler's theorem for convex polyhedra, which states that $f+v=e+2$, where $f$ is the number of faces, $v$ is the number of vertices, and $e$ is the number of edges. This gives $v=e-3$. Since at least three edges emanate from each vertex, $3 v \leq 2 e$, or $v \leq \frac{2 e}{3}$. From these two relationships, $\frac{2 e}{3} \geq e-3$, and thus $e \leq 9$. Considering that the number of vertices is at least 5, since the only 4-vertex solid is clearly the tetrahedron, it follows that $e \geq 8$. Therefore, two cases are possible: $e=8$ or $e=9$. The solution can now be completed similarly to the above.

In this way, it can also be investigated how many types of convex polyhedra with 6 faces exist. The answer is that there are seven.

The solution to the problem is also known for polyhedra with 7, 8, and 9 faces, but not for those with more than 9 faces.