How many triangles can be drawn by means of 9 non-collinear points
A. $84$
B. $72$
C. $144$
D. $126$

Hint: We are given that triangles must be formed from 9 non-collinear points. A triangle has 3 sides. We use the fact that the number of ways of choosing $r$ unordered outcomes from n possibilities is ${}^n{C_r}$ where ${}^n{C_r} = \dfrac{{n!}}{{r!\left( {n - r} \right)!}}$ and $n! = n \times (n - 1) \times (n - 2) \times ...... \times 3 \times 2 \times 1$. Therefore, the number of ways to form a triangle by 9 points is given by ${}^9{C_3}$.

Complete step by step solution:
The number of ways of choosing r unordered outcomes from n possibilities is ${}^n{C_r}$
${}^n{C_r} = \dfrac{{n!}}{{r!\left( {n - r} \right)!}}$
Here n=9 and r=3 since a triangle has 3 sides.
The total number of ways to form a triangle from 9 points are given by, ${}^9{C_3}$
${}^9{C_3} = \dfrac{{9!}}{{3!\left( {9 - 3} \right)!}}$
$ = \dfrac{{9!}}{{3!6!}}$
$ = \dfrac{{9 \times 8 \times 7}}{{3 \times 2}} = 84$

Option ‘A’ is correct

Note: In order to solve the given question, one must know to form and calculate combinations.
The given question can also be solved by using the direct formula to find the number of triangles that can be drawn from n points which is $\dfrac{{n(n - 1)(n - 2)}}{6}$. When n=9 we get the number of triangles to be 84.