144 ??

1 vote

1 vote

Best answer

I have calculated it as 516.

The selection of 3 points out of 16 can be done in 16C3 = 560.

This include points that are collinear. If you calculate the number of collinear lines with *3 grid points then *you will get 44 such lines.

So, final number of possible triangles are 560 -44 = 516

This stackexchange Q&A helped me to solve this: https://math.stackexchange.com/a/634131

0

16C3 is total 3 points selected. -560

10* 4C3 - collinear lines 4 vertical 4 horizontal 2 diagonal -40

I m unable to visualise the other 4 collinear lines

Can u plzz help me out

I think I m missing something...

10* 4C3 - collinear lines 4 vertical 4 horizontal 2 diagonal -40

I m unable to visualise the other 4 collinear lines

Can u plzz help me out

I think I m missing something...

0

For 5* 5 grid

Will it be like this

25C3 all 3 points selected

12* 5C3 = 120 collinear lines 5 each horizontal vertical and 2 diagonal

4 non diagonal collinear lines with 3 grid points

4* 4C3 non diagonal collinear lines with 4 points

I m getting 2160...

But there how they got 12 collinear lines with 3 grid points....:(

Will it be like this

25C3 all 3 points selected

12* 5C3 = 120 collinear lines 5 each horizontal vertical and 2 diagonal

4 non diagonal collinear lines with 3 grid points

4* 4C3 non diagonal collinear lines with 4 points

I m getting 2160...

But there how they got 12 collinear lines with 3 grid points....:(