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Three vertices of a convex n sided polygon are selected. If the number of triangles that can be constructed such that none of the sides of the triangle is also the side of the polygon is 30 then 'n' is

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We proceed to solve this problem in finding the good cases by subtracting the total cases from the "bad" ones:


Total possiblities:

Any combination of $3$ vertices can serve to form a triangle. Number of combinations of $3$ that can be formed from n choices = $\binom{n}{3}$.


Bad case $1$: triangles with exactly $2$ sides in common with the polygon:

Any set of three CONSECUTIVE vertices can serve to form a triangle with exactly $2$ sides in common with the polygon. Thus, given a polygon with $n$ sides, the total number of ways = $n$.


Bad case $2$: triangles with exactly $1$ side in common with the polygon:

Each side can be combined with any NON-ADJACENT vertex to form a triangle with exactly 1 side in common with the polygon. Since there are $n$ sides, each with $n-4$ non-adjacent vertices, the total number of ways = $n(n-4)$.


Thus, the required total is $\binom{n}{3}-n-n(n-4)$. This can be simplified to $$\frac{n(n-4)(n-5)}{6} = 30$$ The answer is thus $\boxed{n = 9}$. Hope it helps.

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Assuming $n\ge 5$:

Total number of triangles: $$\binom n3$$ Triangles such that exactly one of its sides is a side of the polygon: $$n(n-4)$$ Triangles such that exactly two of its sides are sides of the polygon: $$n$$

Then $$\binom n3-n(n-4)-n=30$$

$$n^3-9n^2+20n-180=0$$ $$(n-9)(n^2+20)=0$$

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