There are 10 points in a plane, no three of which are in the same straight line excepting 4, which are collinear. Then, number of

We know that join of any two points gives a line.
The number of lines obtained from 10 points, no three of 
which are collinear ${=}^{10}{\mathrm{C}}_2=\frac{10\times 9}{2\times 1}=45$
Lines obtained from 4 points ${=}^4{\mathrm{C}}_2=\frac{4\times 3}{2\times 1}=6$
∴ Number of lines lost due to 4 collinear points= 6 – 1 = 5
∴ Required number of lines = 45 – 5 = 40.
Also, We know that any triangle can be obtained by joining 
any three points not in the same straight line.
∴ Number of triangles obtained from 10 points, no three 
of which are collinear ${=}^{10}{\mathrm{C}}_3=\frac{10\times 9\times 8}{3\times 2\times 1}=120$
Triangle obtained from 4 points ${=}^4{\mathrm{C}}_3=4$
∴ Number of triangles lost due to 4 collinear points = 4.
∴ Required number of triangles = 120 – 4 = 116.