An equilateral triangle has n equally spaced dots on each side. How many triangles can be formed (of any size)?
Analysis
If there are n dots on each side then total no of dots =n(n+1)/2
Number of combination using 3 at a time =${n(n+1)/2 \choose 3}$
but all set of 3 pts wont be ∆ some will be co linear points
like x1,x2,x3 are coliner on line AB , similarly on CD,EF n so on2
n similarly on other side of ∆ BT, GH & so on
& AT ,SJ & so on
combination of 3 dots on AB = ${n \choose 3}$
on CD = ${n-1 \choose 3}$
on EF = ${n-2 \choose 3}$
so total = ${n \choose 3}+{n-1 \choose 3}+{n-2 \choose 3}....+{3 \choose 3}$
as there are 3 side
$\therefore \ $ total number of such combination
= $3\cdot\left[{n \choose 3}+{n-1 \choose 3}+{n-2 \choose 3}....+{3 \choose 3}\right]$
= $\left[3 \cdot \sum\limits_{k = 3}^n \binom{k}{3} \right]$
$\therefore \ $ total number of triangle
= ${n(n+1)/2 \choose 3}-\left[3 \cdot \sum\limits_{k = 3}^n \binom{k}{3} \right]$
But this is not the answer as x1,x4,x6,x7 are also collinear
also as the length of triangle increases ,we find more co-linearity at different
angles