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.