There are $10$ points in a plane, no three of which are in the same straight line, excepting $4$ points, which are collinear. Find the Number of straight lines obtained from the pairs of these points

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Solution

Since $4$ points are collinear, thus the line formed by these $4$ points will be same.
Therefore,
No. of straight lines $=^{10} C_{2} -^{4} C_{2} + 1 = \frac{10!}{2! \times 8!} - \frac{4!}{2! \times 2!} + 1 = 45 - 6 + 1 = 40$

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There are 10 points in a plane, no three of which are in the same straight line, excepting 4 points, which are collinear. Find the Number of straight lines obtained from the pairs of these points

Since 4points are collinear, thus the line formed by these 4 points will be same.Therefore,No. of straight lines =10C2−4C2+1=10!2!×8!−4!2!×2!+1=45−6+1=40

There are $10$ points in a plane, no three of which are in the same straight line, excepting $4$ points, which are collinear. Find the Number of straight lines obtained from the pairs of these points

Since $4$ points are collinear, thus the line formed by these $4$ points will be same.
Therefore,
No. of straight lines $=^{10} C_{2} -^{4} C_{2} + 1 = \frac{10!}{2! \times 8!} - \frac{4!}{2! \times 2!} + 1 = 45 - 6 + 1 = 40$