Mathematics · Permutations and Combinations

JEE Main 2025 — 3 April, Evening Shift — Question 21

Line $L_{1}$ of slope 2 and line $L_{2}$ of slope $\frac{1}{2}$ intersect at the origin $O$ . In the first quadrant, $P_{1}, P_{2} \ldots P_{12}$ are

12 points on line $L_{1}$ and $Q_{1}, Q_{2} \ldots, Q_{9}$ are 9 points on line $L_{2}$ . Then the total number of triangles, that can

be formed having vertices at three of the 22 points $O, P_{1}, P_{2} \ldots, P_{12}, Q_{1} Q_{2}, \ldots . Q_{9}$ , is:

Answer: C

Total triangles

\text{(2 points as } y = \frac{x}{2}, \text{ 1 point on } y = \frac{x}{2})

+ 2 \text{(points and } y = \frac{x}{2}, \text{ 1 point on } y = 2x)

+ \text{(1 point on } y = 2x, \text{ 1 point on } y = \frac{x}{2} \text{ and origin)}

= {}^9C_2 \cdot {}^{12}C_1 + {}^9C_1 \cdot {}^{12}C_2 + {}^9C_1 \cdot {}^{12}C_1 \cdot {}^1C_1

= 36 \cdot 12 + 9 \cdot 66 + 9 \cdot 12 \cdot 1

= 432 + 594 + 108

= 1134

Answer key and solution verified before publishing.

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