Question

The sides $AB,BC\&CA$ of a triangle $ABC$ have $3,4\&5$ interior points respectively on them.

Find the number of triangles that can be constructed using these interior points as vertices.

• A. $144$
• B. $225$
• C. $205$
• D. None of these

Answer 1

The correct option is C

$205$

Explanation for the correct answer:

Finding the number of triangles that can be constructed using these interior points as vertices:

The total number of points is:$3+4+5=12$

Except for the points that are collinear, all the points will form a triangle.

Therefore, the number of triangles formed is ${}^{12}C_3$

Now, on every side of the triangle, there will be collinear points. Therefore,

$$\begin{gathered} {}^{3}C_3+{}^{4}C_3+{}^{5}C_3\left[{}^{n}C_r=\frac{n!}{r!(n-r)!}\right] \\ \Rightarrow 1+4+10 \\ \Rightarrow 15 \end{gathered}$$

Hence, the required number of triangles is

$$\begin{gathered} {}^{12}C_3–15 \\ \Rightarrow \frac{12!}{3!(12-3)!}-15 \\ \Rightarrow \frac{12!}{3!\times 9!}-15 \\ \Rightarrow \frac{12\times 11\times 10\times 9!}{3\times 2\times 1\times 9!}-15 \\ \Rightarrow \frac{1320}{6}-15 \\ \Rightarrow 220-15 \\ \Rightarrow 205 \end{gathered}$$

Therefore, the correct answer is option (C).