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Ten points are marked on a straight-line and 11 points are marked on another straight-line. How many triangles can be constructed with vertices from among the above points?

  1. 495
  2. 550
  3. 1045
  4. 2475
in Quantitative Aptitude
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Method 1 :

Number of triangles when all (11+10) 21 points are non-colinear = $^{21}C_3$


Number of triangles with 11 and 10 non-colinar points = $^{10}C_3 + ^{11}C_3$ 

 

Thus number of triangles when 11 points colinear and 10 points colinear

= $^{21}C_3$  - ( $^{10}C_3 + ^{11}C_3$ )

=$\frac{21*20*19}{6} - \frac{11*10*9}{6} - \frac{10*9*8}{6}$ = $1045$

 

Method 2:

 

To form a triangle, you have to choose one point from first set and two points from the second set. So, total number of triangles possible: 

($^{10}C_1$)*($^{11}C_2$) + ( $^{10}C_2$)*( $^{11}C_1$) = 1045.

 

Hence,Option(C)1045 is the Answer.

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