JEE Main 2025 Apr 3 Shift 2, Mathematics Q8: Combinations

JEE Main2025Apr 3, Shift 2Mathematics

Line $L_{1}$ of slope $2$ and line $L_{2}$ of slope $\frac{1}{2}$ intersect at the origin $O$ . In the first quadrant, $P_{1}, P_{2}, \dots, P_{12}$ are $12$ points on line $L_{1}$ and $Q_{1}, Q_{2}, \dots, Q_{9}$ 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 $O, P_{1}, P_{2}, \dots, P_{12}, Q_{1}, Q_{2}, \dots, Q_{9}$ , is:

A
$1080$
B
$1134$
C
$1026$
D
$1188$

Solution

A triangle is formed by any three non-collinear points. The only collinear sets among the 22 points are: the 13 points on $L_{1}$ (the origin plus 12 P-points), and the 10 points on $L_{2}$ (the origin plus 9 Q-points).

Count by cases according to where the three vertices lie:

Case 1: 1 vertex from ${P_{i}}$ (12 choices) and 2 from ${Q_{j}}$ ( $(2 9) = 36$ ): $12 \times 36 = 432$ .

Case 2: 2 vertices from ${P_{i}}$ ( $(2 12) = 66$ ) and 1 from ${Q_{j}}$ (9 choices): $66 \times 9 = 594$ .

Case 3: the origin $O$ , plus 1 from ${P_{i}}$ (12 choices) and 1 from ${Q_{j}}$ (9 choices): $12 \times 9 = 108$ .

Total $= 432 + 594 + 108 = 1134$ .

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JEE Main 2025 Apr 3 Shift 2, Mathematics Q8: Combinations

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