Permutations & Combinations
Pentagon from Collinear Points on Triangle Sides
nta_pyq_2026_jan
Grade 11
Question:
Let $ABC$ be a triangle. Consider four points $p_1,p_2,p_3,p_4$ on side $AB$, five points $p_5,p_6,p_7,p_8,p_9$ on side $BC$, and four points $p_{10},p_{11},p_{12},p_{13}$ on side $AC$. None of these points is a vertex of the triangle. Then the total number of pentagons that can be formed by taking all the vertices from the points $p_1,p_2,\ldots,p_{13}$ is _____.
Step-by-Step Solution
Key Concept: For a valid pentagon, no 3 vertices may be collinear, so at most 2 points from each side. Need to choose 5 points with at most 2 from each side (4 on AB, 5 on BC, 4 on AC). Valid distributions $(a,b,c)$ with $a+b+c=5$, $a\leq2$, $b\leq2$, $c\leq2$: permutations of $(1,2,2)$.
Total pentagons $=660$.
Correct Answer: 660