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Question

There are $$8$$ points in a plane out of which $$4$$ are collinear. How many triangles can be formed with these these points as vertices?


A
52
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B
56
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C
48
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D
54
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Solution

The correct option is A $$52$$
There are $$4$$ points collinear. There would be $$3$$ cases
1. Choose any three points from $$4$$ non - collinear points.
2. Choose any $$2$$ points from $$4$$ non collinear and $$1$$ point from collinear points.
3. Choose any $$2$$ points from $$4$$  collinear and $$1$$ point from  non -collinear points.
Total number of ways $$=^4C_3 $$ $$+^4C_1 \times ^4C_2$$ + $$4^C_1 \times ^4C_2$$
$$=52$$

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