Permutations & Combinations
Counting triangles from lines
Grade 11
Question:
<p>In a plane, there are 5 straight lines which pass through a given point, 6 others which all pass through another given point, and 7 others which all pass through a third given point. Supposing no three lines intersect at any point and no two are parallel, find the number of triangles formed by the intersection of the straight line.</p>
Step-by-Step Solution
Key Concept: Triangles are formed by selecting 3 lines from different groups (since lines within the same group are concurrent). Count all possible selections and subtract degenerate cases where all 3 lines don't form a triangle.
<p><strong>Step 1: Understanding the constraint</strong></p><p>We have three groups: 5 lines through point P, 6 lines through point Q, and 7 lines through point R. No three lines are concurrent (except within each group) and no two are parallel.</p><p><strong>Step 2: Valid triangles require non-concurrent lines</strong></p><p>A triangle needs 3 lines with exactly 3 intersection points. Lines from the same concurrent group cannot be used together (they don't create distinct intersection points).</p><p><strong>Step 3: Count total possible selections of 3 lines from 18</strong></p><p>Total ways = C(18,3) = 816</p><p><strong>Step 4: Subtract invalid cases (concurrent triples)</strong></p><p>Concurrent triples from group 1: C(5,3) = 10</p><p>Concurrent triples from group 2: C(6,3) = 20</p><p>Concurrent triples from group 3: C(7,3) = 35</p><p>Total concurrent = 10 + 20 + 35 = 65</p><p><strong>Step 5: Calculate triangles formed</strong></p><p>Number of triangles = 816 - 65 = 751</p><p>∴ Answer: 751</p>
Correct Answer: 751