Given 12 points in a plane, no three of which are collinear. Then number of line segments can be determined, are:

Question

Given 12 points in a plane, no three of which are collinear. Then number of line segments can be determined, are: (A) 76 (B) 66 (C) 60 (D) 80

Tardigrade Admin · Accepted Answer

To find number of line segment we will have to draw the line segments joining two points. If n is the number of such lines segments, then n =12C2 = (12!/2!(12-2)!) = (12 ×11 ×10!/2 ×10!) = 66

Q. Given 12 points in a plane, no three of which are collinear. Then number of line segments can be determined, are:

76

66

60

80

To find number of line segment we will have to draw the line segments joining two points. If n is the number of such lines segments, then
$n =^{12} C_{2} = \frac{12 !}{2 ! ( 12 - 2 ) !} = \frac{12 \times 11 \times 10 !}{2 \times 10 !} = 66$

Q. Given 12 points in a plane, no three of which are collinear. Then number of line segments can be determined, are:

76

66

60

80

To find number of line segment we will have to draw the line segments joining two points. If n is the number of such lines segments, then n=12C2​=2!(12−2)!12!​=2×10!12×11×10!​=66

To find number of line segment we will have to draw the line segments joining two points. If n is the number of such lines segments, then
$n =^{12} C_{2} = \frac{12 !}{2 ! ( 12 - 2 ) !} = \frac{12 \times 11 \times 10 !}{2 \times 10 !} = 66$