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Let $\text{AB, CD, EF, GH}$, and $\text{JK}$ be five diameters of a circle with center at $\text{O}$. In how many ways can three points be chosen out of $\text{A, B, C, D, E, F, G, H, J, K,} $ and $\text{O}$ so as to form a triangle? 

  1. $160$
  2. $159$
  3. $169$
  4. $150$
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Given that, $ \text{AB, CD, EF, GH, and JK}$ be five diameters of a circle with center at $\text{O}.$

We can draw the diagram.



The number of ways in which $3$ points can be selected from $11$ points $= \; ^{11}C_{3} = \frac{11!}{8!\cdot3!} = \frac{11 \cdot 10 \cdot 9 \cdot 8!}{8! \cdot3 \cdot2 \cdot1} = 165$

A straight line can not make a triangle.

In the above diagram, $\text{AOB, COD, EOF, GOH, and JOK}$ cannot make a triangle. We will discard these cases. There are total $5$ combinations.

$\therefore$ The number of ways in which triangle can be formed $= 165-5 = 160.$

Correct Answer $:\text{A}$

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