Number of lines passing through five points such that no three of them are collinear is
10
5
20
8
Idea: One line is decided by any two different points.
No three points are collinear, so every pair of points makes a different line.
We need the number of pairs of points from 5 points:
\(\binom{5}{2}\)
Compute step by step:
\(\binom{5}{2} = \dfrac{5!}{2!\,3!}\)
\(5! = 5 \times 4 \times 3!\)
\(\dfrac{5 \times 4 \times 3!}{2 \times 1 \times 3!}\)
\(\dfrac{5 \times 4}{2 \times 1} = \dfrac{20}{2} = 10\)
Therefore, number of lines = 10. Answer: Option A.