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JEE Main 2024
Permutations & Combinations
Permutations and Combinations
Hard

Question

Line L1L_1 of slope 2 and line L2L_2 of slope 12\frac{1}{2} intersect at the origin O . In the first quadrant, P1\mathrm{P}_1, P2,,P12P_2, \ldots, P_{12} are 12 points on line L1L_1 and Q1,Q2,,Q9Q_1, Q_2, \ldots, Q_9 are 9 points on line L2L_2. Then the total number of triangles, that can be formed having vertices at three of the 22 points O,P1,P2,,P12\mathrm{O}, \mathrm{P}_1, \mathrm{P}_2, \ldots, \mathrm{P}_{12}, Q1,Q2,,Q9\mathrm{Q}_1, \mathrm{Q}_2, \ldots, \mathrm{Q}_9, is:

Options

Solution

Key Concepts and Formulas

  • Combinations: The number of ways to choose rr objects from a set of nn distinct objects, without regard to order, is given by the combination formula: (nr)=n!r!(nr)!{n \choose r} = \frac{n!}{r!(n-r)!}.
  • Triangle Formation: Three points form a triangle if and only if they are not collinear.

Step-by-Step Solution

Step 1: Calculate the total number of ways to choose 3 points from 22 points.

We have a total of 1+12+9=221 + 12 + 9 = 22 points (origin + 12 points on L1L_1 + 9 points on L2L_2). We need to choose 3 points out of these 22 to form a triangle. The total number of ways to choose 3 points is: (223)=22!3!19!=22×21×203×2×1=11×7×20=1540{22 \choose 3} = \frac{22!}{3!19!} = \frac{22 \times 21 \times 20}{3 \times 2 \times 1} = 11 \times 7 \times 20 = 1540 This represents all possible combinations of 3 points.

Step 2: Calculate the number of ways to choose 3 collinear points on L1.

If all three points are chosen from the 13 points on L1L_1 (the origin and the 12 points PiP_i), then they are collinear and do not form a triangle. The number of ways to choose 3 points from these 13 points is: (133)=13!3!10!=13×12×113×2×1=13×2×11=286{13 \choose 3} = \frac{13!}{3!10!} = \frac{13 \times 12 \times 11}{3 \times 2 \times 1} = 13 \times 2 \times 11 = 286

Step 3: Calculate the number of ways to choose 3 collinear points on L2.

Similarly, if all three points are chosen from the 10 points on L2L_2 (the origin and the 9 points QiQ_i), then they are collinear and do not form a triangle. The number of ways to choose 3 points from these 10 points is: (103)=10!3!7!=10×9×83×2×1=10×3×4=120{10 \choose 3} = \frac{10!}{3!7!} = \frac{10 \times 9 \times 8}{3 \times 2 \times 1} = 10 \times 3 \times 4 = 120

Step 4: Calculate the number of triangles that can be formed.

To find the number of triangles, we subtract the number of collinear combinations from the total number of combinations: Number of triangles=(223)(133)(103)=1540286120=1540406=1134\text{Number of triangles} = {22 \choose 3} - {13 \choose 3} - {10 \choose 3} = 1540 - 286 - 120 = 1540 - 406 = 1134

Common Mistakes & Tips

  • Collinear Points: Remember to subtract the cases where the chosen points are collinear, as they do not form triangles.
  • Origin: Include the origin when considering collinear points on L1L_1 and L2L_2.
  • Careful Calculation: Double-check your calculations, especially when dealing with combinations.

Summary

We calculated the total number of ways to choose 3 points from the 22 given points. Then, we subtracted the number of ways to choose 3 points that are collinear on line L1L_1 and line L2L_2. The resulting number represents the total number of triangles that can be formed. This gives us 1134 triangles.

Final Answer

The final answer is \boxed{1134}, which corresponds to option (C).

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