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Permutations and Combinations question

2025 · 8 Apr · Shift 2 · Q41
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Permutations and Combinations question

2025 · 8 Apr · Shift 2 · Q41

JEE MainMathematicsPermutations and CombinationsMCQ+4 / −1
There are 12 points in a plane, no three of which are in the same straight line, except 5 points which are collinear. Then the total number of triangles that can be formed with the vertices at any three of these 12 points is
  1. A
    230
  2. B
    210
  3. C
    200
  4. D
    220
View written solutionFree

Correct answer: B

  1. Total ways to choose any 3 points from 12

Since a triangle is determined by any 3 non-collinear points, first count all possible triples:

(123)=12⋅11⋅103⋅2⋅1=220\binom{12}{3} = \frac{12\cdot 11\cdot 10}{3\cdot 2\cdot 1} = 220(312​)=3⋅2⋅112⋅11⋅10​=220

  1. Subtract the triples that do not form a triangle

We are told that 5 points are collinear. Any choice of 3 points among these 5 will lie on the same straight line, so they will not form a triangle.

Number of such invalid triples:

(53)=5⋅4⋅33⋅2⋅1=10\binom{5}{3} = \frac{5\cdot 4\cdot 3}{3\cdot 2\cdot 1} = 10(35​)=3⋅2⋅15⋅4⋅3​=10

  1. Compute the number of triangles

220−10=210220 - 10 = 210220−10=210

  1. Option check
  • A: 230230230 ❌
  • B: 210210210 ✅
  • C: 200200200 ❌
  • D: 220220220 ❌

Therefore, the total number of triangles is:

210\boxed{210}210​

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