10. The surface of a convex polyhedron is composed of 12 squares, 8 regular hexagons, and 6 regular octagons. At each vertex, one square, one octagon, and one hexagon meet. How many line segments connecting the vertices of the polyhedron lie inside the polyhedron, and not on the faces or edges of the polyhedron?
10. $(840)$

Let $a, b, c, d$ be the number of vertices, edges, face diagonals, and body diagonals, respectively.

Since at each vertex there is one of each type of regular polygon, and the four vertices of each square must be four different vertices of the polyhedron, we have
$c=4 \times$ the number of squares $=4 \times 12=48$.
(The same result can be obtained by considering the hexagons or octagons)
Since three edges emanate from each vertex, and each edge is counted twice, we have,
$$
b=\frac{1}{2} \times 3 a=72 \text{. }
$$

Also, since each square has 2 diagonals, each hexagon has 9 diagonals, and each octagon has 20 diagonals, we have,
$$
c=12 \times 2+8 \times 9+6 \times 20=216 .
$$

Since the polyhedron is convex, we have
$$
d=\mathrm{C}_{48}^{2}-b-c=1128-72-216=840 .
$$