Consider a rectangle ABCD having

Question:

Consider a rectangle $\mathrm{ABCD}$ having $5,7,6,9$ points in the interior of the line segments $\mathrm{AB}, \mathrm{CD}, \mathrm{BC}, \mathrm{DA}$ respectively. Let $\alpha$ be the numberof triangles having these points from different sides as vertices and $\beta$ be the number of quadrilaterals having these points from different sides as vertices. Then $(\beta-\alpha)$ is equal to :

  1. (1) 795

  2. (2) 1173

  3. (3) 1890

  4. (4) 717


Correct Option: , 4

Solution:

$\alpha=$ Number of triangles

$\alpha=5 \cdot 6 \cdot 7+5 \cdot 7 \cdot 9+5 \cdot 6 \cdot 9+6 \cdot 7 \cdot 9$

$=210+315+270+378$

$=1173$

$\beta=$ Number of Quadrilateral

$\beta=5 \cdot 6 \cdot 7 \cdot 9=1890$

$\beta-\alpha=1890-1173=717$

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