Question

The sides AB, BC and AC of a triangle ABC have 3, 4 and 5 interior points on them respectively. Find the number of triangles that can be constructed using these points as vertices.

Answer 1

Hint: Use the fact that the number of triangles formed by joining m non-collinear points is given by $^{m}{{C}_{3}}$. Initially assume that the points are all non-collinear and find the number of triangles formed by them. Then remove the triangles which are formed from joining the collinear points and hence find the number of triangles formed in the above situation.

Complete step-by-step answer:
We know that the number of triangles which can be formed by using m non-collinear points is equivalent to selecting three points out of given m points and is given by $^{m}{{C}_{3}}$.
Now, assume that all the considered points are non-collinear.

Hence, we have 3+4+5=12 non-collinear points.

Hence the number of triangles which can be formed using these points as vertices is $^{12}{{C}_{3}}$.

But in the above calculation we have selected points which are collinear. So, we need to remove those triangles.

The number of triangles which are formed using the points on AB are $^{3}{{C}_{3}}$

The number of triangles which are formed using the points on BC are $^{4}{{C}_{3}}$, and the number of triangles which are formed using the points on AC are $^{5}{{C}_{3}}$.

Hence the number of triangles which can be formed using the points as vertices are $^{12}{{C}_{3}}{{-}^{3}}{{C}_{3}}{{-}^{4}}{{C}_{3}}{{-}^{5}}{{C}_{3}}=\dfrac{12!}{9!3!}-\dfrac{3!}{3!0!}-\dfrac{4!}{3!1!}-\dfrac{5!}{3!2!}=220-1-4-10=205$

Note: [1] Do not try to calculate the number of triangles by forming the diagram of the above situation and counting by hand as the number of triangles is very large.

[2] Many students get confused over when to use combinations and when to use permutations. It can be noted that whenever selections need to be done, we use combinations and whenever arrangements need to be done, we use permutations.