# There are 10 points in a plane out of these 6 are collinear. If N is the number of triangle formed by joining these points. Then

$\begin{array}{1 1}(A)\;N \leq 100\\(B)\;100 < N \leq 140 \\(C)\;140 < N \leq 190 \\(D)\;N > 190 \end{array}$

Number of required triangles $=10C_3-6C_3$
$\qquad= \large\frac{10 \times 9 \times 8}{6} -\frac{6 \times 5 \times 4}{6}$
$\qquad= 120-20$
$\qquad=100$
Hence A is the correct answer.