Mathematics question

Question

On each side of a triangle, 5 points are chosen (other than the vertices of the triangle) and these 15 points are colored red. How many ways are there to choose four red points such that they form the vertices of a quadrilateral?

Answer 1

This is an interesting question! In order for the 4 points to form a quadrilateral, no more than 2 points can be from the same side. So this leaves us with 2 ways we can choose the 4 points: 2 points from 2 different sides OR 1 point from 2 sides and 2 points from the other one. So let's look at both cases and see how many ways we can choose the four points:

2 points each from 2 different sides

₃C₂ x ₅C₂ x ₅C₂ = 300

1. Out of the 3 sides, we are choosing 2.
2. Then for each of those sides - out of the 5 points, we are choosing 2.

1 point from 2 different sides, and 2 points from the third side

₃C₁ x ₅C₁ x ₅C₁ x ₅C₂ = 750

1. Out of the 3 sides, we are choosing 1 (to be our "special" side with 2 points chosen)
2. For two of the sides, we are choosing 1 point out of the 5
3. For the other side, we are choosing 2 points out of the 5

So in total, there are 1050 different ways to choose 4 points such that a quadrilateral is formed.

Let me know if this explanation makes sense or if you need me to elaborate on any details!

Answer 2

i would start by calculating the number of ways one can select 4 points out of 15.

15! / (11! x 4!) = 13 x 14 x 15 / 2 = 1365

Then I would subtract from that number the number of ways one can select 4 points that do not form a quadrilateral,

First let us consider the cases where all 4 points are on the same side. Clearly there are 15 ways to do it.

Now let us consider the remaining cases.

These sets of four points will have three points on the same side, and one point on a different side, For each side there are 4 x 5 / 2 = 10 ways to select 3 points. That gives us 30 ways to select 3 points all of which are on the same side of the triangle. That number should be multiplied by the number of points on the remaining two sides, namely 10. Hence there are 300 ways to select the points.

Finally to answer the question we need to calculate

1365 - 15 - 300 = 1050 ways to select four points which form a quadrilateral