2023 CSAT — Q30

Count all triples of the $6$ points, then subtract triples that are collinear (which fail to form a triangle).

Total triples:

Now check collinearity. The six points are distributed:

• side $AB$: $1$ point,
• side $CD$: $1$ point,
• side $BC$: $2$ points,
• side $DA$: $2$ points.

Three points are collinear only if they lie on one straight line. The most a single side holds is $2$ points (on $BC$ or $DA$) — never $3$. No other line through three of these points exists (a point on $AB$ and a point on $CD$ lie on opposite, parallel sides and are not collinear with a third). So there are zero collinear triples to remove.

Answer: (c) 20.

The triangle count is $\binom{n}{3}$ minus collinear triples — but a line needs three points to be collinear, and here no side carries more than $2$. So the subtraction is zero and the answer is simply $\binom{6}{3} = 20$. The trap is reflexively subtracting “the points on a side,” forgetting that $2$ points can never be a collinear triple. Always verify how many points lie on the same line before deducting.