Question ID: #1167
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:
- (1) 1080
- (2) 1134
- (3) 1026
- (4) 1188
Solution:
$$\text{Total number of triangles} = {}^{9}C_{1}{}^{12}C_{2} + {}^{9}C_{2}{}^{12}C_{1} + {}^{1}C_{1}{}^{9}C_{1}{}^{12}C_{1}$$
$$= 9 \times \frac{12 \times 11}{2} + \frac{9 \times 8}{2} \times 12 + 1 \times 9 \times 12$$
$$= 9 \times 66 + 36 \times 12 + 108$$
$$= 594 + 432 + 108$$
$$= 1134$$
Ans. (2)
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