Question

How many triangles can be formed by joining the vertices of a decagon?

Answer 1

We first describe how triangles are formed and how we can make triangles by joining vertices of a decagon. Then we find the formula of choosing r things out of n things. We use the formula to find the number of ways 3 points can be chosen from the 10 points which is the solution to the problem.

Complete step-by-step solution
A decagon has 10 vertices and 10 sides.
We know that to form a triangle we need three points, which are non-linear.
We also have that no three points out of those 10 vertices are co-linear.

As we are trying to create triangles by joining the vertices of a decagon, we have to choose 3 points out of those 10 points.
So, the number of triangles created by joining the vertices of a decagon is exactly equal to the number of ways 3 points can be chosen from the 10 points.
We know the number of ways “r” things can be chosen out of n things is ${}^{n}{{C}_{r}}=\dfrac{n!}{\left( n-r \right)!\times r!}$.
So, the number of ways 3 points can be chosen from the 10 points is ${}^{10}{{C}_{3}}$.
We find the value of ${}^{10}{{C}_{3}}=\dfrac{10!}{\left( 10-3 \right)!\times 3!}=\dfrac{10!}{7!\times 3!}=\dfrac{10\times 9\times 8}{6}=120$.
The number of triangles created by joining the vertices of a decagon is 120.

Note: We have to remember that in the case of triangles we don’t have to worry but if we are asked about diagonals where we need any 2 points then at the end, we have to subtract the number of sides from the option as they will be considered as diagonals. So, the answer will be ${}^{n}{{C}_{2}}-n$.