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My problem is:

"How many possible line segments can be formed from an $n \times n$ square grid?"

For example by manual counting, if $n = 1$, then the number of line segments $= 0$; if $n = 2$, then the number of line segments $= 6$; and if $n = 3$, then the number of line segments $= 36$. All I can think of is the formula for finding all possible line segments where all points are noncollinear, that is $\frac{n(n -1)}{2}$, but how about a square grid whose intersection of horizontal and vertical line forms a point? How many possible line segments can be formed from such square?

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    $\begingroup$ This tutorial explains how to typeset mathematics on this site. $\endgroup$ – N. F. Taussig Mar 29 at 14:08
  • $\begingroup$ I do not follow. In the $3 \times 3$ case, you have already counted line segments that cross through the center point. $\endgroup$ – N. F. Taussig Mar 29 at 14:10
  • $\begingroup$ No sir, i only count line segments defined by any 2 points on the square grid...but line segments that starts from opposite ends of the square that cross the center is not counted... $\endgroup$ – rosa Mar 29 at 14:19
  • $\begingroup$ I only get 28 for 3 points across $\endgroup$ – Roddy MacPhee Mar 29 at 16:40
  • $\begingroup$ I got 28 by not counting those line segments crossing another point/s I misunderstood... so if p = 1, l = 0, p = 2, l = 6, if p = 3 l = 28... is there any pattern evident? Im suck at combinatorics thats why I tried manual counting then derive formula algebraically... $\endgroup$ – rosa Mar 30 at 10:55

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