GGUY wrote:
S is a set of points in the plane. How many distinct triangles can be drawn that have three of the points in S as vertices?
(1) The number of distinct points in S is 5.
(2) No three of the points in S are collinear.
We are given that S is a set of points in the plane and we must determine how many distinct triangles can be drawn with three of the points in S as vertices. So essentially, we must determine how many distinct triangles can be drawn with the points provided.
Statement One Alone: The number of distinct points in S is 5.
Using the information in statement one, it may be tempting to conclude that the number of triangles that can be drawn is 5C3 = (5 x 4 x 3)/3! = 10 triangles. However, because we do not know the positioning of the points, we cannot actually say that 10 distinct triangles can be created. Let’s say, for instance, that all the points were collinear, which means that they are all located on one line. If that were the case, we would not be able to create any triangles. Thus, statement one is not sufficient to answer the question. We can eliminate answer choices A and D.
Note: We were able to determine that 10 triangles could be formed with 5 points if and only if no 3 points are collinear. Only then would the number of triangles be 5C3 = 10.
Statement Two Alone: No three of the points in S are collinear.
Using the information in statement two, we cannot answer the question because we do not know how many points are in S. We can eliminate answer choice B.
Statements One and Two Together: Using the information from statements one and two we know that we have 5 points in the plane and that no three points are collinear. Thus, we can determine that the number of triangles that can be created in the plane is 5C3 = 10.
Answer: C
Hi Scott, I understand the logic behind the answer and got C as my answer. However, why does 5C3 work? I don´t understand that concept and maybe it will come in the exame in a PS type.