# How many triangles can be formed from the 12 non-collinear points?

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

• It should depend on whether given any three points, they are collinear or not. e.g Let all the points be on circumference of a circle. Then the answer is C(11,2). – NeerajKumar Oct 7 '15 at 11:00
• @NeerajKumar, can you move your comment to answer? So I can choose and upvote your answer. Thanks – akusaja Oct 7 '15 at 12:44
• Well, it's quite unnecessary, if you got the answer. :) – NeerajKumar Oct 7 '15 at 13:18
• @NeerajKumar: No, it most certainly is not “unnecessary”. Unanswered questions only clog up the Unanswered Questions Queue, which is why one of the rules of this site is to never answer a question in the comment section. After posting your answer, ask the OP to accept it $($not just upvote it, but accept it$)$. – Lucian Oct 7 '15 at 14:15
• @Lucian : Alright buddy. Point taken. I just meant that this was a routine exercise in combinatorics so it, most probably, wont be missed, even if removed since it gets reduced to the benefit of asker only. Though looking at the site in its entirety, your point is entirely relevant too. – NeerajKumar Oct 7 '15 at 14:19