Question

If the sides $AB,BC,$ and $CA$ of a triangle $ABC$ have $3$, $5$ and $6$ interior points respectively, then the total number of triangles that can be constructed using these points as vertices is equal to:

• A. $360$
• B. $240$
• C. $333$
• D. $364$

Answer 1

The correct option is C

$333$

Explanation of correct answer:

Permutation :

We can construct a triangle by choosing any one point of all three sides.

We can also construct a triangle by choosing one point from one side and remaining two points from either sides.

Number of ways of choosing $1$points out of $3$ points from the side $AB={}^3\mathrm{ℂ}_1$

Number of ways of choosing $1$points out of $5$ points from the side $BC={}^5\mathrm{ℂ}_1$

Number of ways of choosing $1$points out of $6$ points from the side $AC={}^6\mathrm{ℂ}_1$

Number of ways of choosing $2$ points out of $6$ points from the side $AC={}^6\mathrm{ℂ}_2$

Number of ways of choosing $2$ points out of $5$ points from the side $BC={}^5\mathrm{ℂ}_2$

Number of ways of choosing $2$ points out of $3$ points from the side $AB={}^3\mathrm{ℂ}_2$

Therefore total number of triangles will be,

$={}^{3}C_1\times {}^{5}C_1\times {}^{6}C_1+{}^{3}C_1\times {}^{5}C_2+{}^{5}C_1\times {}^{3}C_2+{}^{3}C_1\times {}^{6}C_2+{}^{6}C_1\times {}^{3}C_2+{}^5\mathrm{C}_1\times {}^6\mathrm{C}_2+{}^6\mathrm{C}_1\times {}^5\mathrm{C}_2$

$$\begin{gathered} =90+30+15+45+18+75+60 \\ =333 \end{gathered}$$

Hence, option (C) is the correct answer.