Probability of getting an obtuse triangle when choosing three points on a circle. The problem is: $1153$ points are evenly distributed on a circle. Find the probability of randomly picking $3$ points constituted vertices of an obtuse triangle.
The given solution says
$$
\frac{\binom{1153}{1}\binom{576}{2}}{\binom{1153}{3}} = \frac{1725}{2302}
$$
What I don't understand is how we get $\binom{576}{2}$. Where does this come from? I am assuming that $576$ comes from $1153-1 \over 2$ but I cant figure out why we do that, and why we have to choose 2 points from them. Any help?
 A: The triangle will be obtuse when all three vertices are within the same half of the circle. We can always designate one vertex as the "start" of a half circle, moving clockwise. Choose one point to be the vertex at the start of this half circle; there are $1153 \choose 1$ ways to choose this point. There are 1152 points remaining, half of which are eligible for the other two points on the same half as the first point in the clockwise direction; so there are ${1152/2 \choose 2}={576 \choose 2}$ ways to choose the other two vertices.
A: The number $576\choose 2$ comes from the number of ways to choose the other two vertices of the obtuse-angled triangle, once you've chosen the vertex with the obtuse angle.
With the point chosen for your obtuse angle, label all points with numbers $-576$ to $576$, e.g. in clockwise order, so that your chosen point is labelled as $0$. We need to pick two of those - one at each side of the given point $0$ (i.e. one negative and one positive), so that the angle is actually obtuse. This means that the "distance" of the second and the third point is at most $576$.
One can see that, if you pick $-576$, you have no choices for the third point at all (as you cannot choose $0$ again, and any positive point is "further away"). If you pick $-575$ you can only pick $1$. If you pick $-574$, you can also pick $1$ or $2$ etc. until, if you pick $-1$, you can pick any of the the points $1,2,3,\ldots,575$.
So the total number of ways to pick the "other" two points is $1+2+3+\ldots+575={576\choose 2}$.
I suggest you test this logic on something simpler, i.e. instead of $1153$ points choose any small odd number of points (say $5$ or $7$) and convince yourselves that the same logic works. E.g. for $5$ points there is only one way to choose the other two ponts, for $7$ points there are $1+2=3$ ways etc.

Update: I can see from @Golden_Ratio's answer that the value $576\choose 2$ pops out directly if, instead of the obtuse angle vertex we choose the acute angle vertex (the first in, say, clockwise, order). Still, I will keep this answer as a potentially different (less efficient for sure!) solution.
A: Angle subtended by adjacent points at centre = $2\pi/1153$
Let points be marked $1, 2, .. 1153$.
Let us pick $1$ as first point. We cannot pick both the remaining points from the set $289,290,... 865$. That leaves us with $576$ points to choose the two from.
One could argue that why is there a need to pick both points from the set $289, 290, .. 865$. We can just pick one. But this scenario will be included when a point in this set $289, ... 865$ is the vertex (or the first point). So we don't double count.
