In a plane there are 14 points, of which 4 are collinear and the remaining are in general position. How many distinct triangles can be formed using these 14 points as vertices?

Difficulty: Easy

Correct Answer: 360

Explanation:

Introduction / Context:The number of triangles from a set of points is normally C(n, 3), but any collinear triple cannot form a triangle. We adjust by subtracting those degenerate triples.

Given Data / Assumptions:

  • Total points n = 14.
  • Exactly 4 are collinear; every other triple is non-collinear.

Concept / Approach:Valid triangles = total triples − collinear triples. Only the special set of 4 contributes collinear triples.

Step-by-Step Solution:Total triples: C(14, 3) = 364.Collinear triples among the 4: C(4, 3) = 4.Triangles = 364 − 4 = 360.

Verification / Alternative check:If no three were collinear, C(14, 3) would be correct. One collinear block reduces the count by exactly its C(4, 3) degenerate triples.

Why Other Options Are Wrong:364 ignores collinearity; 368 and 365 exceed the possible total; 356 subtracts too many.

Common Pitfalls:Subtracting C(4,2) instead of C(4,3) or double-counting.

Final Answer:360

More Questions from Permutation and Combination

Discussion & Comments

No comments yet. Be the first to comment!
Join Discussion