Diagonal Dilemma

Here's the 36 sided polyhedron with 12 squares and 24 triangles, with 26 vertices. See Khulbe's solution for figuring out the number of space diagonals.

In order to solve this problem, there are two formulas that must be applied. it is that the E(edges) = (F(Faces) * S(sides))/2 and that F + V(vertices) = E + 2. So we know that there are 12 + 24 = 36 faces. then using the formula E = (F S)/2, we get (12 * 4(the amount of sides a square has) + 24 3(amount of sides triangle has))/2 = 60. So our figure has 60 edges. By using the other formula, F + V = E + 2, we manipulate it to get V = E + 2 - F = 60 + 2 - 36 = 26. So our figure has 26 vertices. Now to find out all the possible diagonals. We can have 26 C 2 = 26!/(2!*24!) = 325 diagonals( We used combinatorics because out of the 26 vertices, we can pick 2). Lastly all we need to do is subtract all the impossible space diagonals from 325. We know that all the edges are not space diagonals and that all the diagonals in the squares which is 12 *2 since there are 2 diagonals per square( triangles don't have diagonals) are not space diagonals. So we subtract 60 + 24 = 84 from 325 = 241.