Number of triangles formed by joining the vertices of n sided polygon, which has no side common with that of the polygon, is

Question

Number of triangles formed by joining the vertices of n sided polygon, which has no side common with that of the polygon, is (A) (n(n-3)/2) (B) ((n - 4)(n - 5)/3!) (C) (n(n - 4)(n - 5)/3!) (D) None of these

Tardigrade Admin · Accepted Answer

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)

∴

Number of triangles having no side common with that of polygon

=

(Total Number of triangles i.e.

^{n} C_{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_{1} A_{2} ..... A_{n})

then three consecutive vertices can be chosen as like

A_{1} A_{2} A_{3}, A_{2} A_{3} A_{4}, ...., A_{n} A_{1} A_{2}

.

∴

Required triangles having no side common with that of the polygon

=^{n} C_{3} - n (n - 4) - n

= \frac{n}{6} [n^{2} - 9 n + 20] = \frac{n ( n - 4 ) ( n - 5 )}{3 !}

Q. Number of triangles formed by joining the vertices of $n$ sided polygon, which has no side common with that of the polygon, is

$\frac{n ( n - 3 )}{2}$

$\frac{( n - 4 ) ( n - 5 )}{3 !}$

$\frac{n ( n - 4 ) ( n - 5 )}{3 !}$

None of these

Q. Number of triangles formed by joining the vertices of n sided polygon, which has no side common with that of the polygon, is

2n(n−3)​

3!(n−4)(n−5)​

3!n(n−4)(n−5)​

None of these

Q. Number of triangles formed by joining the vertices of $n$ sided polygon, which has no side common with that of the polygon, is

$\frac{n ( n - 3 )}{2}$

$\frac{( n - 4 ) ( n - 5 )}{3 !}$

$\frac{n ( n - 4 ) ( n - 5 )}{3 !}$