I am currently self-studying introductory combinatorics, and I don't fully get an example in the book.

The question was as follows: If no three diagonals of a convex decagon meet at the same point, into how many line segments are the diagonals divided by their intersections?.

So I understand that the total number of intersections would be $\binom {10} {4} = 210$ since for every 4 vertices there will be 1 intersection, and also the number of diagonals would be $10 \choose 2$-10 = 35. I also understand that the number of line segments are $k+1$ when there are $k$ intersections along a line. I however don't get the answer the author gave which was $35+2\times210$. Why were the number of diagonals added?, why did he multiply the number of intersections by 2?, shouldn't it be 4 since for every intersection there will be 4 segments?. If anyone can explain this to me, I'd be very grateful. Also if someone can provide a different way to solve this problem, that would be awesome.


You start with 35 diagonals. Each intersection point adds a segment to both of the intersecting diagonals. Therefore answer is 35 + twice number of intersections.

  • $\begingroup$ I can't believe it was that simple, thank you very much. $\endgroup$ – turingcomplete Aug 1 '12 at 7:42

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.