I feel really bad for asking so many questions... but just one more... there are 12 points that are equally spaced on a circle. Define a set of points $(A,B,C,D)$ NICE if $AB$ intersect $CD$. How many nice sets are there?
What I tries to do is counting in cases... there isn't any case when they are just one point away, and there is $1\cdot 9$ case when two points away , $2\cdot 8$ 3 points away until $5\cdot 5$ when 6 points away. there are 12 points in every cases and there are $2^3$ ways to arrange $A,B,C,D$. Wrapping this up, we will have:$$8\cdot12\cdot(1\cdot9+2\cdot8+3\cdot7+4\cdot6+5\cdot5)=9120$$
However this is no way near the answer. Can anyone tell where i did it wrong?