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Permutations and Combinations question

2021 · 16 Mar · Shift 2 · Q28
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Permutations and Combinations question

2021 · 16 Mar · Shift 2 · Q28

JEE MainMathematicsPermutations and CombinationsMCQ+4 / −1
Consider a rectangle ABCD having 5, 7, 6, 9 points in the interior of the line segments AB, CD, BC, DA respectively. Let α\alphaα be the number of triangles having these points from different sides as vertices and β\betaβ be the number of quadrilaterals having these points from different sides as vertices. Then (β−α\beta-\alphaβ−α) is equal to :
  1. A
    717
  2. B
    795
  3. C
    1890
  4. D
    1173
View written solutionFree

Correct answer: A

Let the numbers of interior points on the sides be:

  • On ABABAB: 555
  • On BCBCBC: 666
  • On CDCDCD: 777
  • On DADADA: 999

We must count:

  1. α\alphaα = number of triangles whose vertices are chosen from different sides.
  2. β\betaβ = number of quadrilaterals whose vertices are chosen from different sides.

1. Counting α\alphaα

A triangle is formed by choosing points from three different sides of the rectangle.

So we choose any 3 sides out of 4, and then one point from each chosen side.

The four possible side-triples are:

  1. AB,BC,CDAB, BC, CDAB,BC,CD:
    5⋅6⋅7=2105 \cdot 6 \cdot 7 = 2105⋅6⋅7=210

  2. AB,BC,DAAB, BC, DAAB,BC,DA:
    5⋅6⋅9=2705 \cdot 6 \cdot 9 = 2705⋅6⋅9=270

  3. AB,CD,DAAB, CD, DAAB,CD,DA:
    5⋅7⋅9=3155 \cdot 7 \cdot 9 = 3155⋅7⋅9=315

  4. BC,CD,DABC, CD, DABC,CD,DA:
    6⋅7⋅9=3786 \cdot 7 \cdot 9 = 3786⋅7⋅9=378

Hence,

α=210+270+315+378=1173\alpha = 210+270+315+378 = 1173α=210+270+315+378=1173

2. Counting β\betaβ

A quadrilateral having vertices from different sides means we choose one point from each of the four sides.

Thus,

β=5⋅6⋅7⋅9\beta = 5 \cdot 6 \cdot 7 \cdot 9β=5⋅6⋅7⋅9

Now,

5⋅6=30,7⋅9=635\cdot 6 = 30, \quad 7\cdot 9 = 635⋅6=30,7⋅9=63

so

β=30⋅63=1890\beta = 30 \cdot 63 = 1890β=30⋅63=1890

3. Compute β−α\beta-\alphaβ−α

β−α=1890−1173=717\beta-\alpha = 1890 - 1173 = 717β−α=1890−1173=717

4. Option check

  • A: 717717717 ✅
  • B: 795795795
  • C: 189018901890
  • D: 117311731173

Therefore, the correct answer is:

717\boxed{717}717​

So, Option A is correct.

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