Question

How many triangles can be formed by joining the vertices of an octagon?
(a) 56
(b) 28
(c) 112
(d) 120

Answer 1

Hint: We start solving the problem by recalling the fact that the octagon has 8 vertices and 8 sides. We then draw the figure and check whether there are any collinear points in it. We then use the fact that the number of ways of choosing ‘r’ items out of ‘n’ items $\left( n\ge r \right)$ is ${}^{n}{{C}_{r}}$ ways for choosing 3 points out of the 8 vertices. We then make use of the properties ${}^{n}{{C}_{r}}=\dfrac{n!}{r!\left( n-r \right)!}$ and $n!=n\times \left( n-1 \right)\times \left( n-2 \right)\times ......\times 3\times 2\times 1$ to get the required number of triangles.

Complete step-by-step solution
According to the problem, we are given that we need to find the number of triangles that can be formed by joining the vertices of the triangle.
We know that the octagon is a figure with 8 vertices and 8 sides. Now, let us draw the octagon first.

From the figure, we can see that no three points are collinear. We know that a triangle can be formed only if we have three non-collinear points.
Now, we need to find ways to choose three points in order to form a triangle as we know only one triangle can be formed with a fixed three points.
We know that the number of ways of choosing ‘r’ items out of ‘n’ items $\left( n\ge r \right)$ is ${}^{n}{{C}_{r}}$ ways. Using this we get the number of ways to choose 3 points out of 8 vertices will be ${}^{8}{{C}_{3}}$ ways.
We know that ${}^{n}{{C}_{r}}=\dfrac{n!}{r!\left( n-r \right)!}$ and $n!=n\times \left( n-1 \right)\times \left( n-2 \right)\times ......\times 3\times 2\times 1$.
So, we have ${}^{8}{{C}_{3}}=\dfrac{8!}{3!\left( 8-3 \right)!}=\dfrac{8!}{3!5!}$.
$\Rightarrow {}^{8}{{C}_{3}}=\dfrac{8\times 7\times 6}{3\times 2\times 1}$.
$\Rightarrow {}^{8}{{C}_{3}}=56$ ways.
We have found that the number of triangles that can be formed by joining the vertices of an octagon is 56.
$\therefore$ The correct option for the given problem is (a).

Note: We should not know that there should not be any collinear points while forming a triangle. Whenever we get this type of problem, we first try to recall the number of vertices present in it and then check whether there can be any collinear points in it. We have taken regular octagon for making the calculation easier and avoiding calculations. Similarly, we can expect problems to find the number of triangles and diagonals that can be formed with the vertices of the concave octagon.