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
- A \(449\)
- B \(419\)
- C \(455\)
- D \(443\)
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 \(={ }^{15} C_3-12=455-12=443\)
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