JEE Mains · Maths · STD 11 - 6. permutation and combination
Line \(L_1\) of slope 2 and line \(L_2\) of slope \(\frac{1}{2}\) intersect at the origin O . In the first quadrant, \(\mathrm{P}_1, \mathrm{P}_2, \ldots . \mathrm{P}_{12}\) 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 \(\mathrm{O}, \mathrm{P}_1, \mathrm{P}_2, \ldots \mathrm{P}_{12}\), \(\mathrm{Q}_1, \mathrm{Q}_2, \ldots . \mathrm{Q}_9\), is:
- A \(1080\)
- B \(1134\)
- C \(1026\)
- D \(1188\)
Answer & Solution
Correct Answer
(B) \(1134\)
Step-by-step Solution
Detailed explanation
Total number of \(\Delta\) are \(\begin{aligned} & ={ }^9 \mathrm{C}_1{ }^{12} \mathrm{C}_2+{ }^9 \mathrm{C}_2{ }^{12} \mathrm{C}_1+{ }^1 \mathrm{C}_1{ }^9 \mathrm{C}_1{ }^{12} \mathrm{C}_1 \\ & =594+432+108 \\ & =1134 \end{aligned}\)
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