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AP EAMCET · Maths · Permutation Combination

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

  1. A \(449\)
  2. B \(419\)
  3. C \(455\)
  4. D \(443\)
Verified Solution

Answer & Solution

Correct Answer

(D) \(443\)

Step-by-step Solution

Detailed explanation

Total number of distinct triangles \(={ }^{15} C_3\) Now, we have to exclude that cases in which \(i+j+k=15\) Total number of cases in which \(i+j+k=15\) is 12 . \(\therefore\) Total number of required triangles \(={ }^{15} C_3-12=455-12=443\)