Counting on a circle 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?

I’m assuming that you’re actually counting ordered $4$-tuples $\langle a,b,c,d\rangle$ of distinct points (chosen from the $12$) such that $ab$ intersects $cd$.
Let the points be $p_1,p_2,\ldots,p_{12}$ in order around the circle. Let $\{i,j,k,\ell\}$ be a $4$-element subset of $\{1,2,\ldots,12\}$, where $i<j<k<\ell$. Then $p_ip_k$ intersects $p_jp_\ell$. To get a $4$-tuple $\langle a,b,c,d\rangle$ from these $4$ points in such a way that $ab$ intersects $cd$, we can let $a$ be any one of the $4$. The choice of $a$ completely determines that of $b$: if $a=p_i$, $b$ must be $p_k$ and vice versa, and if $a=p_j$, $b$ must be $p_\ell$ and vice versa. We then have $2$ possible choice for $c$, after which $d$ must be the remaining point. Thus, there are $4\cdot2=8$ possible $4$-tuples using $p_i,p_j,p_k$, and $p_\ell$. Since there are $\binom{12}4=495$ ways to choose $4$ of the $12$ points, there are $495\cdot8=3960$ $4$-tuples $\langle a,b,c,d\rangle$ such that $ab$ intersects $cd$.
