Question

triangles can be formed in an octagon by connecting any 3 of its vertices. determine the number of different triangles that can be formed in an octagon.

Explanation:

Step1: Identify the combination formula

The number of combinations of choosing $r$ elements from $n$ elements is given by $C(n,r)=\frac{n!}{r!(n - r)!}$. Here, $n$ is the number of vertices of the octagon ($n = 8$) and $r$ is the number of vertices needed to form a triangle ($r=3$).

Step2: Substitute values into the formula

$C(8,3)=\frac{8!}{3!(8 - 3)!}=\frac{8!}{3!5!}$.
Since $n!=n\times(n - 1)\times\cdots\times1$, we have $8! = 8\times7\times6\times5!$, so $C(8,3)=\frac{8\times7\times6\times5!}{3!5!}$.
And $3!=3\times2\times1 = 6$.
Then $C(8,3)=\frac{8\times7\times6}{6}=56$.

Answer:

56