# How many quadrilaterals are there whose vertices are the vertices of the 17-gon and have 2 sides that are diagonals?

I started by using the formula to find the number of diagonals, and then did that number choose 2, since if I pick the two diagonals, the other two sides are determined. How do I remove the vertices that cross each other?

I'm interpreting the requirement "$2$ sides are diagonals" as: "exactly two sides of the quadrilateral $Q$ are diagonals (i.e., non-edges) of the $17$-gon $P$". This then implies that exactly two special sides of $Q$ are edges of $P$. It is easier to work with this condition. Denote the vertices of $P$ by $0$, $1$, $\ldots$, $16$.
The two special sides of $Q$ can be adjacent in $Q$, and then they will also be adjacent in $P$. There are $17$ ways to choose the hinge vertex on $P$. Assume we have chosen vertex $0$. Then vertices $16$, $0$, and $1$ are vertices of $Q$. Vertices $15$ and $2$ will then be forbidden for $Q$, since these would lead to a third "special" side. This leaves the $12$ vertices $3$, $4$, $\ldots$, $14$ of $P$ as allowed fourth vertex of $Q$.
The two special sides of $Q$ can be opposite in $Q$. There are $17$ ways to choose a first edge in $P$. With a similar counting as above we then find that there are $12$ admissible ways to choose a second edge in $P$ which is separated by two true diagonals from the first chosen edge.
In all we obtain $17\cdot 12+{1\over2}\cdot 17\cdot 12=306$ admissible quadrilaterals.