Hi! I was hoping that someone could help me with this question. I know how to solve it intuitively, but I was reading through Vince and Brian’s PowerPrep Solutions and got a little confused. They explain how to solve this question using combinatorics. The apply the combinations formula (picture of their explanation below). However, I’m not sure where this formula comes from? I though the combinations formula was n!/r! (n-r)!. If someone could help me with this problem I would greatly appreciate it. Thank you!
There’s not much “combination” happening tbh.
For quantity A, if you fix one point then you see that you can have two diagonals (from our fixed point) which satisfy the requirement (parallel to at least one side of octagon). Because of the rotational symmetry of our regular octagon, we know that this holds for any arbitrary point we pick.
Thus, we have: 8(2) = 16 diagonals following the aforementioned logic. However, since a diagonal is a line segment with two endpoints, we have instances of double counting in our total, and so the actual number of diagonals would be \frac{16}{2} = 8.
For Quantity B, can you conjecture a formula to count the number of diagonals in an n-sided polygon? If so, Quantity B is just the total number of diagonals in an octagon - Quantity A.
I think it’s fairly obvious which quantity is bigger from here on out.
Thank you! I find your method easier to understand. I was not sure where combinatorics was coming from.