A [convex](https://artofproblemsolving.com/wiki/index.php/Convex) [polyhedron](https://artofproblemsolving.com/wiki/index.php/Polyhedron) has for its [faces](https://artofproblemsolving.com/wiki/index.php/Face) 12 [squares](https://artofproblemsolving.com/wiki/index.php/Square), 8 [regular](https://artofproblemsolving.com/wiki/index.php/Regular_polygon) [hexagons](https://artofproblemsolving.com/wiki/index.php/Hexagon), and 6 regular [octagons](https://artofproblemsolving.com/wiki/index.php/Octagon). At each [vertex](https://artofproblemsolving.com/wiki/index.php/Vertex) of the polyhedron one square, one hexagon, and one octagon meet. How many [segments](https://artofproblemsolving.com/wiki/index.php/Segment) joining vertices of the polyhedron lie in the interior of the polyhedron rather than along an [edge](https://artofproblemsolving.com/wiki/index.php/Edge) or a [face](https://artofproblemsolving.com/wiki/index.php/Face)?
The polyhedron described looks like this, a truncated cuboctahedron.

[Truncatedcuboctahedron.jpg](https://artofproblemsolving.com/wiki/index.php/File:Truncatedcuboctahedron.jpg) [](https://artofproblemsolving.com/wiki/index.php/File:Truncatedcuboctahedron.jpg)
The number of segments joining the vertices of the polyhedron is ${48\choose2} = 1128$. We must now subtract out those segments that lie along an edge or a face.
Since every vertex of the polyhedron lies on exactly one vertex of a square/hexagon/octagon, we have that $V = 12 \cdot 4 = 8 \cdot 6 = 6 \cdot 8 = 48$.
Each vertex is formed by the intersection of 3 edges. Since every edge is counted twice, once at each of its endpoints, the number of edges $E$ is $\frac{3}{2}V = 72$.
Each of the segments lying on a face of the polyhedron must be a diagonal of that face. Each square contributes $\frac{n(n-3)}{2} = 2$ diagonals, each hexagon $9$, and each octagon $20$. The number of diagonals is thus $2 \cdot 12 + 9 \cdot 8 + 20 \cdot 6 = 216$.
Subtracting, we get that the number of space diagonals is $1128 - 72 - 216 = \boxed{840}$.