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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?

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    Uh i think i over-counted it... but i am too brain-dead to figure over-counts so can anyone just tell me the solution... – 2017-01-02
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    Are you counting *sets* of $4$ points or *ordered* $4$-*tuples* $\langle a,b,c,d\rangle$ such that $ab$ intersects $cd$? – 2017-01-02

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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

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    oh i see! so any 4 element from the set makes 2 perpendicular line and we just need to count the ways that they intersect. Thanks ! – 2017-01-02
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    @alex: Not literally perpendicular in general, but definitely intersecting, yes. You’re welcome! – 2017-01-02
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I initially got your same answer and my way of thinking was same as that of you. Then I got the error, we cannot just multiply with $2^3$ to take care of order of $4$ points in circle as intersection lines may differ. Corrected count must be as given below.

$2 \times 12 \times\left[\dbinom{1}{1}\dbinom{9}{1}+\dbinom{2}{1}\dbinom{8}{1}+\dbinom{3}{1}\dbinom{7}{1}+\dbinom{4}{1}\dbinom{6}{1}+\dbinom{5}{1}\dbinom{5}{1}+\dbinom{6}{1}\dbinom{4}{1}+\dbinom{7}{1}\dbinom{3}{1}+\dbinom{8}{1}\dbinom{2}{1}+\dbinom{9}{1}\dbinom{1}{1}\right]\\=3960$

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    Oh yeah! what is your motive behind this? Does it focus on counting stuff that will not cause overcount? I always use the wrong way when solving counting stuff... – 2017-01-03