We dont have to. Because these selections are already removed when calculating 9C3.hitmis wrote:Why should we not have to also subtract combinations of (x1,y1) = (x2,y2)=(x3,y3) , 9 such combinations, which would also not create a triangle.?
Evaluate 9C3
9C3=9!*8!*7!/(3!*2!*1!)
Here, 8and7 shows reduced choices. Hence, repeated choices are already removed.