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Permutations and Combinations question

2011 · Shift 0 · Q47
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Permutations and Combinations question

2011 · Shift 0 · Q47

JEE MainMathematicsPermutations and CombinationsMCQ+4 / −1
These are 10 points in a plane, out of these 6 are collinear, if N is the number of triangles formed by joining these points. then:
  1. A
    N≤100N \le 100N≤100
  2. B
    100<N≤140100 \lt N \le 140100<N≤140
  3. C
    140<N≤190 140 \lt N \le 190\,140<N≤190
  4. D
    N>190N \gt 190N>190
View written solutionFree

Correct answer: A

  1. Total triangles from 10 points

If no three points were collinear, the number of triangles formed would be (103)=120.\binom{10}{3} = 120.(310​)=120.

  1. Subtract degenerate selections

We are told that 6 points are collinear. Any selection of 3 points from these 6 will not form a triangle.

Number of such degenerate triples: (63)=20.\binom{6}{3} = 20.(36​)=20.

  1. Actual number of triangles

Hence, N=(103)−(63)=120−20=100.N = \binom{10}{3} - \binom{6}{3} = 120 - 20 = 100.N=(310​)−(36​)=120−20=100.

  1. Check the options
  • A: N≤100N \le 100N≤100 → true, since N=100N=100N=100
  • B: 100<N≤140100 < N \le 140100<N≤140 → false
  • C: 140<N≤190140 < N \le 190140<N≤190 → false
  • D: N>190N > 190N>190 → false

Therefore, the correct option is: A\boxed{A}A​

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