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Combination Based on Geometry

The sides AB ...

Question

The sides $A B, B C$ and $C A$ of $△ A B C$ are marked with $3, 4$ and $5$ interior points respectively. Number of triangles that can be constructed using these interior points as vertices is

205

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150

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200

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210

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Solution

The correct option is A $205$
We have total $3 + 4 + 5 = 12$ points.
So, number of $△ s$ that can be formed using $12$ such points $=$ Total number of ways of selecting $3$ points $-$ Number of ways $3$ collinear points are selected
$=^{12} C_{3} -^{3} C_{3} -^{4} C_{3} -^{5} C_{3} = 205$

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Combination Based on Geometry

Standard XII Mathematics

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The sides AB, BC and CA of △ABC are marked with 3, 4 and 5 interior points respectively. Number of triangles that can be constructed using these interior points as vertices is

The correct option is A 205We have total 3+4+5=12 points. So, number of △s that can be formed using 12 such points = Total number of ways of selecting 3 points − Number of ways 3 collinear points are selected = 12C3− 3C3− 4C3− 5C3=205

The sides $A B, B C$ and $C A$ of $△ A B C$ are marked with $3, 4$ and $5$ interior points respectively. Number of triangles that can be constructed using these interior points as vertices is

The correct option is A $205$
We have total $3 + 4 + 5 = 12$ points.
So, number of $△ s$ that can be formed using $12$ such points $=$ Total number of ways of selecting $3$ points $-$ Number of ways $3$ collinear points are selected
$=^{12} C_{3} -^{3} C_{3} -^{4} C_{3} -^{5} C_{3} = 205$