I'm trying to figure out how the formula to calculate the number of different combinations of polygons that could coexist in a circle.
For instance, let's say I have a circle with one nail every $5$ degrees ($72$ nails). I'm using rubber bands to generate perfect triangles and squares, and I want each nail to get at most one single rubber band:
To calculate how many different combinations of one square and one equilateral triangle I can draw in the canvas, I would think:
A circle with $72$ nails contains a vertex of the very same square every $72/4 = 18$ nails.
The very same way, it contains a vertex of an equilateral triangle every $72/3 = 24$ nails.
This means that I will have $18$ different possible squares with first vertex on each of those $18$ nails (different squares that leverage on the very same set of nails are counted only once).
For each square, I could draw $24$ different triangles (same thing: different triangles whose vertexes falls on the very same set of nails are counted only once), but that would not account for triangles whose vertexes falls over a nail already occupied by a square.
Because the vertexes of a square occupy $4$ nails in the circle, I should not count any triangle that would fall on any of those $4$ nails… which leads me to think the number of triangles for each circle would be $24-4=20$ bringing the total combinations to $18 \cdot (24-4)=360$.
This looks solid when reverted: I draw a triangle first, it occupies $3$ vertexes, I can add $18-3$ different squares bringing the total to $24 \cdot (18-3)=360$. Good.
Let's now consider drawing $2$ squares and $2$ triangles on the same circle:
The first square has $18$ different starting vertexes, the second square has $17$ (I will have $\dfrac{18 \cdot 17}{2} =153$ different unique combinations of $2$ squares).
Let's add the triangles: the first triangle has availability of $24-(2 \cdot 4)=16$ nails [$2 \cdot 4$ nails already occupied by the squares] and the second one has $15$ available nails.
I'd be led to think I can have a total of $\dfrac{(18 \cdot 17)(22-6)(21-6)}{2 \cdot 2}=18360$ possible combinations of $2$ squares and $2$ triangles on my circle.
But the solution doesn't work because just by rewriting the formula considering the triangles first leads to a completely different result:
Two triangles occupy $3 \cdot 2=6$ nails, so there could be $\dfrac{24 \cdot 23 \cdot (18-6)(17-6)}{2 \cdot 2} =18216$ combinations.
The very same formula results in $18360$ and $18216$ for the same scenario… so it's not a very reliable one. 🙂
Any help would be very much appreciated.
Best Answer
I considered rotating the triangle with the square keeping fixed. Basically you have to rotate the triangle till its one vertex reaches the point just before its second last vertex. In this process whenever any vertex of the triangle is coinciding with any vertex of the square, you don't consider that combo. However if you do the same for multiple different figures, you find that when you are not considering a combination, then one vertex of that triangle might be touching one square, and another vertex of he triangle might be touching another square. Considering two squares and 1 triangle, start rotating the triangle assuming the two squares to be a single figure. Then change the square combinations and do the same rotation with the triangle...