There are 12 distinct non-collinear points in a same plane, they are points A,B,….L.
How many different triangle can be formed, with criteria one of its vertice must be contain point A?
My attempt:
Because the arrangements didn't need an order, so we can use combination to solve this problem.
Since there are 12 points and we need only to take three of them, so the possibility is
C(12,3) = 220
If there is no criteria, I think this is the total number to create the triangle from the 12 distinct points.
But, how about the numbers of solution if the criteria is required one of its vertice must be point A? Is the total possibility remains the same?
Thanks
Best Answer
It should depend on whether given any three points, they are collinear or not.( Your question is a bit fuzzy about what is exactly the meaning of collinearity. Is it that all the points are not on a line or that no three points in the set in on a line)
If no three points in the set are collinear, then similarly we can choose any two points other than A to be other two vertices of a triangle. Hence the answer is C(11,2).