A [convex](https://artofproblemsolving.com/wiki/index.php/Convex) polyhedron $P$ has $26$ vertices, $60$ edges, and $36$ faces, $24$ of which are triangular and $12$ of which are quadrilaterals. A space diagonal is a line segment connecting two non-adjacent vertices that do not belong to the same face. How many space diagonals does $P$ have?
Every pair of [ vertices](https://artofproblemsolving.com/wiki/index.php/Vertex) of the [polyhedron](https://artofproblemsolving.com/wiki/index.php/Polyhedron) determines either an [edge](https://artofproblemsolving.com/wiki/index.php/Edge), a face [diagonal](https://artofproblemsolving.com/wiki/index.php/Diagonal) or a space diagonal.  We have ${26 \choose 2} = \frac{26\cdot25}2 = 325$ total [line segments](https://artofproblemsolving.com/wiki/index.php/Line_segment) determined by the vertices.  Of these, $60$ are edges.  Each [triangular](https://artofproblemsolving.com/wiki/index.php/Triangle) face has $0$ face diagonals and each [quadrilateral](https://artofproblemsolving.com/wiki/index.php/Quadrilateral) face has $2$, so there are $2 \cdot 12 = 24$ face diagonals.  This leaves $325 - 60 - 24 = \boxed{241}$ segments to be the space diagonals.