Question

Choose 3, 4, 5 points other than vertices respectively on the sides AB, BC and CA of a \[\Delta {\text{ABC}}\]. The number of triangles that can be formed by using only these points as vertices, is
A. 220
B. 217
C. 215
D. 210
E. 205

Answer 1

Hint: To solve this question we will use the formula ${}^{\text{n}}{{\text{C}}_{\text{r}}}$ = $\dfrac{{{\text{n!}}}}{{{\text{r!(n - r)!}}}}$ to find the number of combinations of vertices to form a triangle. A triangle is formed from 3 vertices.

Complete step-by-step solution:

Now, we have 3 points on side AB, 4 points on side BC, 5 points on side CA. Now, we have to exclude the vertices A, B and C.
Now, total number of vertices = 3 + 4 + 5 = 12 vertices.
So, total number of triangles that can be formed from these vertices = ${}^{12}{{\text{C}}_3}$
Now, ${}^{12}{{\text{C}}_3}$ also contains the combination which includes all three points from the same side. All points on the same side are collinear and a triangle can’t be made from collinear points. So, we have to remove that case from ${}^{12}{{\text{C}}_3}$. Also, two points from the same side can contribute in a triangle.
So, total number of triangles formed = ${}^{12}{{\text{C}}_3}{\text{ - }}{}^3{{\text{C}}_3}{\text{ - }}{}^4{{\text{C}}_3}{\text{ - }}{}^5{{\text{C}}_3}$ = $\dfrac{{{\text{12!}}}}{{{\text{3!9!}}}}{\text{ - }}\dfrac{{{\text{3!}}}}{{{\text{3!0!}}}}{\text{ - }}\dfrac{{{\text{4!}}}}{{{\text{3!1!}}}}{\text{ - }}\dfrac{{{\text{5!}}}}{{{\text{3!2!}}}}$
Therefore, total number of triangles = 220 – 1 – 4 – 10 = 205
So, option (E) is correct.

Note: The given question can be solved by another method. In that method, we have to take cases to solve the given problem. Such a method is also correct but it takes more time and there are more chances of error. So, applying the formula is the easiest method to solve such types of questions.