Line, Circle or a Cross?

Quick answer: any three of the ten points in arrangement B form a triangle (there are $\binom{10}{3}$ of these).

There are also ten points in arrangement A, but this time some choices of three points don't form a triangle.

In arrangement C, there are just nine points, and not every choice of three works, so there are definitely fewer triangles.

So arrangement B has the most triangles.

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Slow answer: to make a triangle in arrangement A, we need two points in one row and one in the other. We get a total of $2\cdot \binom{5}{2} \cdot \binom{5}{1}=100$ different triangles.

In arrangement B, as above, there are $\binom{10}{3}=120$

In arrangement C, there are a total of $\binom{9}{3}=84$ ways to choose three points. But $\binom{5}{3}=10$ of these lie on the vertical line, and another $10$ on the horizontal. So there are $\binom{9}{3}-2\cdot \binom{5}{3}=64$ triangles in arrangement C.

There is no restriction for choosing the three dots in circle, hence it can form the most non-degenerate triangles.

Figure A has 5 dots each in both the lines. To form a non-degenerate triangle, we could either have the 2 dots on the first line and the remaining dot on the second line or vice versa. So choosing 2 dots from any line first is basically choosing 2 objects from 5 objects in no order i.e $\large { _{ 5 }{ C }_{ 2 } }=10$ and then we could choose 1 dot out of the 5 from the other line, so total triangles is $10 \times 5=50$ but as there are 2 lines, we could do first with the up line and then with down so there are twice more triangles actually. Thus arrangement A can form $50 \times 2=100$ triangles.

For figure B as all points are not colinear, we choose any 3 points with no order which is choosing 3 dots out of the total 10 dots, i.e $\large { _{ 10 }{ C }_{ 3 } }=120$ total triangles.

For figure C, the figure has the same no. of dots in the vertical and horizontal line, if we exclude the middle dot, we could first choose the 2 dots from the horizontal line i.e out of 4 dots which is $\large { _{ 4 }{ C }_{ 2 } }=6$ and then we could choose the remaining 1 out of the 4 in the vertical line, so there are total $6 \times 4=24$ triangles, however we could take the 2 dots from the vertical line and then the remaining from the horizontal so there are twice as much, $24 \times 2=48$ triangles. Now our first dot could be the middle one, and we could choose any 1 out of the 4 from the vertical line and one out of the 4 from the horizontal line, so $4 \times 4=16$ triangles. So total triangles = 16+48=64.

As 120>100>64, Arrangement B can form the most triangles.