Question

Let \(P_1,P_2,\ldots\ldots,P_{15}\) be 15 points on a circle. The number of distinct triangles formed by points \(P_i,P_j,P_k\) such that \(i+j+k\ne15\), is :

• A. \(12\)
• B. \(419\)
• C. \(443\)
• D. \(455\)

Hint:

Subtract from \(\binom{15}{3}\) the number of triples of distinct indices with \(i+j+k=15\).

Answer 1

The Correct Option is C

Any 3 of the 15 points on a circle form a triangle: \(\binom{15}{3}=455\).
Count triples \(i\lt j\lt k\) of distinct positive integers with \(i+j+k=15\): \(i=1\): \((2,12),(3,11),(4,10),(5,9),(6,8)\) = 5; \(i=2\): \((3,10),(4,9),(5,8),(6,7)\) = 4; \(i=3\): \((4,8),(5,7)\) = 2; \(i=4\): \((5,6)\) = 1; \(i\ge5\): none.
Total \(=12\).
Required number \(=455-12=443\).
Answer: (3) 443