Over the past few weeks, I have covered several tricky math concepts. Below is a problem that incorporates many of these lessons into one problem. If you are up to it, definitely take a stab at this tough, time-consuming problem. If you want a little primer (or already gave it a crack, but are unsure how to proceed), have a look here at combinations and permutations.
How many unique quadrilaterals can be inscribed in the vertices of a nonagon (a 9-sided figure), if points A and B, two vertices in the nonagon, cannot make up the same quadrilateral?
How to Crack It
We know that any four out of nine points can make a quadrilateral. If we approach this problem as a combinations question, then we want to figure out how many ways we can select 4 out of 9 points. This will yield the total. From this, we want to subtract those points that contain both the letters A and B.
First, though, let’s find the total number of quadrilaterals: (9 x 8 x 7 x 6)/(4 x 3 x 2 x 1) = 3 x 6 x 7 = 126.
Next, and here’s the tricky part, we have to figure out how many quadrilaterals contain both points A and B. You could write down those points: ABCD, ABCE, etc. But that, of course, would take forever, and give rise to writer’s cramp faster than it will yield an answer.
Instead, let’s look at this logically: two points are already determined, A and B. There are 7 remaining points that could occupy the two other vertices of the quadrilateral (we can think of it as ABXX – we have to choose two out of the seven letters to fit in the XXs). Therefore, using combinations, we get (7 x 6)/(2 x 1) = 21.
That is, there are 21 quadrilaterals that contain points A and B. We have to discount these if we want to answer the question. Therefore, we take the total, 126 – 21 = 105 (B).
This is obviously a very difficult problem: it is conceptually tough, and you have to use embedded combinations (that just sounds scary). But, hopefully, after watching the videos and/or reading the posts on the various concepts, you can combine the different elements necessary to solve this problem.
Again, as with all challenge problems, don’t despair, unless you positively have to break 700.