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?


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    $\begingroup$ 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). $\endgroup$ Oct 7, 2015 at 11:00
  • $\begingroup$ @NeerajKumar, can you move your comment to answer? So I can choose and upvote your answer. Thanks $\endgroup$
    – akusaja
    Oct 7, 2015 at 12:44
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    $\begingroup$ Well, it's quite unnecessary, if you got the answer. :) $\endgroup$ Oct 7, 2015 at 13:18
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    $\begingroup$ @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$)$. $\endgroup$
    – Lucian
    Oct 7, 2015 at 14:15
  • $\begingroup$ @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. $\endgroup$ Oct 7, 2015 at 14:19

1 Answer 1


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).

  • 3
    $\begingroup$ This +1 is not just for the answer, but also the positive attitude shown in your earlier comments $\endgroup$
    – Shailesh
    Oct 7, 2015 at 15:34

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