IPMAT Indore 2022 · Quantitative Ability · Question 4

Step 1 — note what makes a triangle. Three points form a triangle only if they are not collinear. Since all 7 points lie on one line and all 5 on a parallel line, any three chosen from the same line are collinear and give no triangle.

So every valid triangle takes 2 points from one line and 1 from the other.

Step 2 — count both cases.

2 from the line of 7, 1 from the line of 5:

\left( \genfrac{}{}{0pt}{}{7}{2}\right)\times 5=21\times 5=105

2 from the line of 5, 1 from the line of 7:

\left( \genfrac{}{}{0pt}{}{5}{2}\right)\times 7=10\times 7=70

Step 3 — add.

105+70=175

Answer: 175

Shortcut. Complement check, if you prefer it: total ways to pick 3 of the 12 points minus the collinear ones,

\left( \genfrac{}{}{0pt}{}{12}{3}\right)-\left( \genfrac{}{}{0pt}{}{7}{3}\right)-\left( \genfrac{}{}{0pt}{}{5}{3}\right)=220-35-10=175

Both routes are quick; the complement is safer when there are more than two lines, because you never have to enumerate the split.

The one thing to get right is that the two lines are parallel, so no three points can be collinear across lines — which is exactly why the 2+1 split is always valid.