2013 AMC 10A Problem 25

If no three diagonals were concurrent, each choice of $4$ vertices would determine one interior intersection, giving $\binom84=70$. The $4$ long diagonals through opposite vertices all meet at the center, so the center was counted $\binom42=6$ times and should be counted once. Subtract $5$. There are also $8$ symmetric points where $3$ diagonals meet. Each was counted $\binom32=3$ times and should be counted once, so subtract $8(3-1)=16$. The number of distinct interior intersection points is $70-5-16=49$. Thus, A is the correct answer.