First slide

Combinations

Question

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

Easy

A

straight lines formed by joining them is 40
B

triangles formed by joining them is 116
C

straight lines formed by joining them is 45
D

triangles formed by joining them is 120

Solution

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 $=^{10} C_{2} = \frac{10 \times 9}{2 \times 1} = 45$
Lines obtained from 4 points $=^{4} C_{2} = \frac{4 \times 3}{2 \times 1} = 6$
∴ 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 $=^{10} C_{3} = \frac{10 \times 9 \times 8}{3 \times 2 \times 1} = 120$
Triangle obtained from 4 points $=^{4} C_{3} = 4$
∴ Number of triangles lost due to 4 collinear points = 4.
∴ Required number of triangles = 120 – 4 = 116.

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