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There are n sides of a polygon(where $n>5$).

Triangles are formed by joining the vertices of the polygon. How many triangles can be constructed with no side common to the polygon?

My try:

Total possible triangles = $\frac{(n)(n-1)(n-2)}{6} = \binom{n}{3}$ ------(1)

Triangles with 2 sides common = $n$ ------(2)

Triangles with 1 side common = $n(n-4)$ ------(3)

So, with no side common = $1-2-3$

Is there any other way to get it directly without following this process?

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We count the number of "good" triangles with one vertex painted blue.

The blue vertex can be chosen in $n$ ways. For every one of these ways, the two neighbouring vertices are forbidden. That leaves $n-3$ vertices. Write down $n-5$ stars, with a little space between them, like this: $$\ast\qquad\ast\qquad\ast\qquad\ast\qquad\ast\qquad\ast\qquad\ast\qquad\ast$$ That determines $n-4$ "gaps" ($n-5$ ordinary gaps, plus $2$ endgaps) to slip our $2$ remaining vertices into.

There are $\binom{n-4}{2}$ ways to choose these gaps. So there is a total of $n\binom{n-4}{2}$ good triangles with a blue vertex.

This counts each uncoloured good triangle $3$ times. So the number of these is $\frac{1}{3}\cdot n\binom{n-4}{2}$.

Note that the idea generalizes to good (convex) quadrilaterals, and so on.

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First let us find the number of solutions to $$d_1+d_2+d_3=n$$ such that $d_i\geq 2$. This is same as the number of solutions in positive integers to $$d_1+d_2+d_3=n-3$$ By stars and bars method this is $\binom{n-4}{2}$. For each such solution, we have $n$ possible triangles and each triangle is counted three times. Thus the number of triangles is $\frac{n}{3}\binom{n-4}{2}$

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Suppose you choose a corner of the polygon(n possibilities). Then for the next vertex, you can choose any vertex that is not within one unit of the first vertex (n-3 possibilities). Then proceed with casework:

Case 1: The second vertex is within 2 units of the first vertex. (2 possibilities) Then there are 5 points you cannot choose for the the next vertex. So the total from this case is $n(2)(n-5)=2n^2-10n$

Case 2: any other vertex. The total from this case is $n(n-3-2)(n-6)=n^3-11n^2+30n.$ Therefore the total is $n^3-9n^2+20n=n(n-4)(n-5)$.

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Looking clockwise, attach one unused vertex $\Large\bullet$ to each of the $3$ used vertices $\Large\circ$ that will form the triangle, viz. $\boxed{\Large{\circ\bullet}}$

Now there are $3$ boxes + $(n-6) = (n-3)$ objects.

Place the boxes in $\binom{n-3}{3}$ ways,
but you are allowing objects only $(n-3)$ places instead of $n$, so multiply by $\frac{n}{n-3}$ to get

formula $=\frac{n}{n-3}\times\binom{n-3}{3}$

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