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Q.
Number of triangles formed by joining the vertices of $n$ sided polygon, which has no side common with that of the polygon, is
Permutations and Combinations
Solution:
Total number of triangles formed
= Triangle having no side common + triangle having exactly one side common + triangles having exactly two sides common (with those of the polygon)
$\therefore $ Number of triangles having no side common with that of polygon
$= $ (Total Number of triangles i.e. $\,{}^nC_3) -$ Number of triangles having exactly one side common - Number of triangles having exactly two sides common.
Now, number of triangles having exactly one side common
$= n(n - 4)$
and for the number of triangles having exactly two side common, we have to select three consecutive vertices of the polygon. If vertices of polygon are marked by $(A_1A_2.....A_n)$ then three consecutive vertices can be chosen as like $A_1A_2A_3, A_2A_3A_4, ...., A_nA_1A_2$.
$\therefore $ Required triangles having no side common with that
of the polygon $=\,{}^nC_3 - n(n - 4) - n$
$ = \frac{n}{6}[ n^2 - 9n + 20 ] = \frac{n(n-4)(n - 5)}{3!}$