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Option 3 : 10
Maximum number of intersection points = Number of pairs of lines that can be made
⇒ 5C2 = 5!/ (3! × 2!) = 5 × 4/2 = 10
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Answered
Option 3 : 10
Maximum number of intersection points = Number of pairs of lines that can be made
⇒ 5C2 = 5!/ (3! × 2!) = 5 × 4/2 = 10