non-isosceles triangles with vertices in a 20-sided regular polygon. Suppose $A_1A_2 . . . A_{20}$ is a $20-$sided regular polygon. 
How many non-isosceles (scalene) triangles
can be formed whose vertices are among the vertices of the polygon but whose sides are not
the sides of the polygon?
I could'nt find a cute answer to this problem. My answer is different from those at other websites. Not to mention that the answers are also different at different websites. 
 A: We first count all nondegenerate triangles containing no edge $A_iA_{i+1}$, and then subtract  the isosceles triangles among these.
Put the first vertex $v_1$ at $A_0=A_{20}$. If you put $v_2$ at $A_2$ or $A_{18}$ then there are $15$ allowed $A_i$ left for $v_3$. If you put $v_2$ at an $A_i$ with $3\leq i\leq 17$  then there are $2\cdot 3=6$ forbidden $A_i$ for $v_3$. It follows that you can choose $(v_1,v_2,v_3)$ in
$$20\cdot2\cdot 15+20\cdot 15\cdot 14=4800$$
ways, giving rise to ${1\over6}\cdot4800=800$ different triangles.
The tip of an isoscles triangle can be chosen in $20$ ways, and then the base in $8$ ways. Since there is no equilateral triangle possible  it follows that there are $160$ isosceles triangles.
The total number of admissible triangles  therefore is $640$.
A: Given a regular polygon of $2n$ sides, 
start by counting all the distinct triangles that can be formed
from the vertices of a $2n$-gon:
$
\binom{2n}{3}.
$
Then eliminate the ones that don't belong to the set by counting these
disjoint sets of triangles:


*

*Isoceles triangles. There are $n - 1$ different possible lengths for each of the equal sides, and $2n$ possible placements of the vertex between those sides, so $2n(n-1)$ such triangles if $n$ is not a multiple of $3$. (If $n=3k$ then the preceding procedure counts each equilateral triangle three times and we must remove the duplicates. Compare this answer.)

*Triangles with exactly one side on the $2n$-gon. Note that no such triangle can be isoceles. There are $2n$ ways to choose the side on the $2n$-gon, and the third vertex can be any of the $2n-4$ points not adjacent to the vertices of the common side, so there are $2n(2n-4)$ such triangles.


There is no need for a third set of triangles consisting of those that share two sides in common with the $2n$-gon; those were included among the isoceles triangles.
The total number of triangles in all three excluded sets is
$2n(n - 1) + 2n(2n-4) = 2n(3n-5)$.
So the total number of scalene triangles not sharing a side with
the $2n$-gon is
$$
\binom{2n}{3} - 2n(3n-5).
$$
Setting $2n = 20$, we have 
$$\binom{2n}{3} - 2n(3n-5) = \binom{20}{3} - 20(25) = 1140 - 500 = 640.$$
A: Total number of $\triangle  =  $no. of $\triangle$ with no side common $+$ no. of $\triangle$ with one side common $+$ no. of $\triangle$ with two sides are common
$\bullet\;$ Triangle with one side common $\displaystyle = \binom{n-4}{1} \times n$
$\bullet\;$ Triangle with two sides are common $\displaystyle  = n$ 
So no. of $\triangle$ with no side common $$ = \binom{n}{3}-n(n-4)-n$$ 
Put $n=20$
A: Let $a<b<c$ be the "sidelengths" of the triangle. The given conditions then enforce
$$a=2+x_1,\quad b=3+x_1+x_2,\quad c=4+x_1+x_2+x_3$$
with
$$x_i\geq 0\quad(1\leq i\leq 3),\qquad 3x_1+2x_2+x_3=11\ .\tag{1}$$
Choosing $x_1:=0, 1, 2, 3$ in turn produces $6+5+3+2=16$ solutions of $(1)$. Each of the $16$ resulting shapes can be placed in $40$ ways, hence there are $640$ admissible triangles in all.
A: No. of scalene triangles 
$$= C_{20}^3-180= \frac{20\cdot 19\cdot 18}{6} - 180 = 960$$
