Question:

Let \(P_1,P_2,\ldots,P_{15}\) be \(15\) points on a circle. The number of distinct triangles formed by points \(P_i,P_j,P_k\) such that \(i+j+k\neq 15\), is

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When a condition excludes some combinations, first count the total number of selections and then subtract the cases that violate the condition.
Updated On: Jun 26, 2026
  • \(449\)
  • \(419\)
  • \(455\)
  • \(443\)
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The Correct Option is D

Solution and Explanation

Step 1: Find the total number of triangles.
There are \(15\) points on a circle.
Since any \(3\) distinct points on a circle form a triangle, the total number of triangles is \[ {}^{15}C_3 \] \[ =\frac{15\cdot14\cdot13}{3\cdot2\cdot1} \] \[ =455 \]

Step 2: Count the triangles for which \(i+j+k=15\).
We need distinct positive integers \[ i\lt j\lt k \] such that \[ i+j+k=15. \] The possible triples are \[ (1,2,12),\ (1,3,11),\ (1,4,10),\ (1,5,9),\ (1,6,8) \] \[ (2,3,10),\ (2,4,9),\ (2,5,8),\ (2,6,7) \] \[ (3,4,8),\ (3,5,7),\ (4,5,6) \] Thus, the number of triples satisfying \[ i+j+k=15 \] is \[ 12. \]

Step 3: Subtract the unwanted triangles.
Required number of triangles is \[ 455-12 \] \[ =443. \]

Step 4: Final conclusion.
Therefore, the required number of distinct triangles is \[ \boxed{443} \]
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