grehater123456 wrote:
why can this not be treated like a round table problem? for the hexagons, we fix one point and the remaining 5 points can be arranged in the following ways: 1*9*8*7*6*5. For the quadrilaterals, we have 1*9*8*7. So the value of the ratio will be a fraction and B will be greater
I have found an answer to this, in case anyone else was curious. In circular questions, we fix a point and treat it as a permutations when the order matters. This is applicable to round table arrangements where we use (n-1)! by fixing a point. However, in this question, the order of choosing the points does not matter -- no matter which of the 4 points we choose first in a quadrilateral, we get the same quadrilateral. So we do not fix points here and treat it as a combination.