Ten points lie in a plane such that no three | vedclass

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(B) Let the $10$ points be arranged as vertices of a convex decagon. $A$ line passing through two points divides the remaining $8$ points into two set

Vedclass · Accepted Answer

$5$

Vedclass · Answer

(B) Let the $10$ points be arranged as vertices of a convex decagon. $A$ line passing through two points divides the remaining $8$ points into two sets of $4$ points each if and only if the line is a diagonal that skips $4$ vertices on one side and $4$ vertices on the other.In a convex decagon with vertices labeled $1, 2, \dots, 10$ in order,such a line connects vertex $i$ to vertex $i+5$ (where indices are taken modulo $10$).The possible lines are $(1, 6), (2, 7), (3, 8), (4, 9), 	ext{ and } (5, 10)$.Thus,there are exactly $5$ such lines.

Ten points lie in a plane such that no three of them are collinear. The number of lines passing through exactly two of these points and dividing the plane into two regions,each containing four of the remaining points,is

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